Welcome all to the first ever Brilliant Inequality Contest. Like the Brilliant Integration Contest, the aim of the Inequality Contest is to improve skills and techniques often used in Olympiad-style Inequality problems. But above all, the main reason is so we can have fun.

Anyone is allowed to participate, as long as they adhere to the following rules.

I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

Please make a substantial comment. Spam or unrelated comments will be deleted. (To help me out, try to delete your comments within an hour of when you posted them, if they are not the solutions.)

Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 48 hours. Then, you must post the solution and you have the right to post a new problem.

If the one who solves the last problem does not post a new problem in 24 hours, the creator of the previous problem has the right to post another problem.

The scope of the problems is Olympiad-style inequalities.

You are not allowed to post problems requiring calculus in the solutions (use of differentiation to prove a curve is concave or convex is allowed).

Lagrange Multipliers are not allowed to be used in a solution.

Inequalities allowed to be used are AM-GM, Muirhead, Power Mean and Weighted Power Mean, Cauchy-Schwarz, Holder, Rearrangement, Chebyshev, Schur, Jensen, Karamata, Reverse Rearrangement and Titu's lemma.

Try to post the simplest solution possible. For example, if someone posted a solution using Holder, Titu's and Cauchy, when there is a solution using only AM-GM, the latter is preferred.

Format your proof as follows:

SOLUTION OF PROBLEM(insert problem number here)

[Post your solution here]

PROBLEM(insert problem number here)

[Post your problem here]

Remember to reshare this note so it goes to everyone out there. And above all else, have fun!!!

PROBLEM 1Let \(x\), \(y\) and \(z\) be positive reals such that \(x+y \geq z\), \(y+z \geq x\) and \(z+x \geq y\). Find families of solutions for \((x, y, z)\) such that the following inequality is satisfied.

\[2x^2 (y+z) + 2y^2 (z+x) + 2z^2 (x+y) \geq x^3 + y^3 + z^3 + 9xyz\]

*P.S.*: For those who want to discuss problem solutions, they can do so here.

## Comments

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TopNewestSolution of Problem 1 :We will show that the given inequality is true for all positive reals \(x,y,z\) satisfying the conditions given in the question.

Case 1:When x,y,zaresides of triangle.Let \(\color{Blue}{x=a+b,y=b+c,z=a+c} \); \(\text{(a,b,c>0)} \),so our inequality becomes:

\[\color{Purple}{\sum_{cyc}2(a+b)^2(a+b+2c) \geq \sum_{cyc} (a+b)^3 +9(a+b)(b+c)(a+c)}\]

\[\implies 2 \sum_{cyc}(a+b)^{3} + 4\sum_{cyc} c(a+b)^{2} \geq \sum_{cyc}(a+b)^3 +9(a+b)(b+c)(a+c) \]

\[\implies \sum_{cyc} (a+b)^{3} +4\sum_{cyc} c(a+b)^{2} \geq 9(a+b)(b+c)(a+c) \]

\[\implies 2 \sum_{cyc}a^{3}+ 3\sum_{cyc} (ab^{2} +ba^{2}) + 4\sum_{cyc} (ab^{2} +ba^{2}) + 24abc \geq 9(a+b)(b+c)(a+c) \]

\[\implies 2 \sum_{cyc} a^{3} + 7(a+b)(b+c)(c+a) + 10abc \geq 9(a+b)(b+c)(c+a)\]

\[\implies 2 \sum_{cyc} a^{3}+ 10abc \geq 2(a+b)(b+c)(c+a) \]

\[\implies \sum_{cyc} a^{3}+ 5abc \geq (a+b)(b+c)(c+a) \]

\[\color{Red}{\implies \sum_{cyc} a^{3} + 3abc \geq a^{2} b + b^{2} a + a^{2} c + c^{2} a + b^{2} c + c^{2} b}\]

which is true by Schur's Inequality.

Equality occurs when \(a=b=c\) , that is \(x=y=z\).

Hence proved.

Case 2:When x,y,zare notsides of a triangle.In this case either \(x+y=z\) or \(y+z = x\) or \(z+x=y\).

Let's take \(\color{Blue}{x+y =z} \).

Since the inequality is symmetric, WLOG we can assume that \(\color{Green}{z \ge x \ge y}\).

Now substituting \(x+y = z\) in the inequality , we get \[\color{Red}{ 2x^{2}(2y+x) + 2y^{2}(2x+y) + 2(x+y)^{3} \geq x^{3} + y^{3} + (x+y)^{3} + 9xyz}\]

After rearranging the expression, we get,\[ x^{3} + y^{3} \geq x^{2} y + y^{2}x\]

which is true by Rearrangement Inequality

Hence proved. – Harsh Shrivastava · 1 year, 4 months ago

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– Sharky Kesa · 1 year, 3 months ago

Can you write the equality cases for Case 2?Log in to reply

– Harsh Shrivastava · 1 year, 3 months ago

I don't think equality exists . . .Log in to reply

Problem 8For positive reals \(a,b,c\), prove that \[\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\le \dfrac{1}{abc}\] – Daniel Liu · 1 year, 4 months ago

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Solution to Problem 8I used a massive expansion, but I've triple-checked so its right (If you haven't figured it out, expansion is what I do with inequalities). We have to prove

\[\left (\dfrac {1}{a+b} + \dfrac{1}{b+c} \right ) \left (\dfrac {1}{b+c} + \dfrac{1}{c+a} \right ) \left (\dfrac {1}{c+a} + \dfrac{1}{a+b} \right ) \leq \dfrac {1}{abc}\]

\[\left (\dfrac {a+2b+c}{(a+b)(b+c)} \right ) \left (\dfrac {b+2c+a}{(b+c)(c+a)} \right ) \left (\dfrac {c+2a+b}{(c+a)(a+b)} \right ) \leq \dfrac {1}{abc}\]

Simplifying the denominators, we have

\[abc(a+2b+c)(b+2c+a)(c+2a+b) \leq (a+b)^2 (b+c)^2 (c+a)^2\]

Simplifying all the brackets, we get

\[\begin{align} &2a^4bc + 2ab^4c + 2abc^4 + 7a^3b^2c + 7a^3bc^2 + 7a^2b^3c + 7ab^3c^2 + 7a^2bc^3 + 7ab^2c^3 + 16a^2b^2c^2 \leq\\ &2a^4bc + 2ab^4c + 2abc^4 + 6a^3b^2c + 6a^3bc^2 + 6a^2b^3c + 6ab^3c^2 + 6a^2bc^3 + 6ab^2c^3 + 10a^2b^2c^2 \\& \quad \quad \quad \quad \quad+a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2 + 2a^3b^3 + 2b^3c^3 + 2c^3a^3 \end{align}\]

Simplifying, we get

\[a^3b^2c + a^3bc^2 + a^2b^3c + ab^3c^2 + a^2bc^3 + ab^2c^3 + 6a^2b^2c^2 \leq a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2+2a^3b^3+2b^3c^3+2c^3a^3\]

By AM - GM, we have

\[\dfrac {a^4c^2 + a^2b^4}{2} \geq a^3b^2c, \quad \dfrac {a^4b^2 + a^2c^4}{2} \geq a^3bc^2, \quad \dfrac {b^4c^2 + b^2a^4}{2} \geq a^2b^3c\]

\[\dfrac {b^4a^2 + b^2c^4}{2} \geq ab^3c^2, \quad \dfrac {c^4b^2 + c^2a^4}{2} \geq a^2bc^3, \quad \dfrac {c^4a^2 + c^2b^4}{2} \geq ab^2c^3\]

\[a^3b^3 + b^3c^3 + c^3a^3 \geq 3a^2b^2c^2 \quad \Rightarrow \quad 2a^3b^3 + 2b^3c^3 + 2c^3a^3 \geq 6a^2b^2c^2\]

Summing these, we get the expansion above. Thus, the statement is true. – Sharky Kesa · 1 year, 3 months ago

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We just need to prove \[\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\left(\dfrac{c}{b+c}+\dfrac{c}{c+a}\right)\left(\dfrac{a}{c+a}+\dfrac{a}{a+b}\right)\le 1\] However by AM-GM \[LHS\le \left(\begin{array}{c}\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{c+a}+\dfrac{a}{a+b}\\ \hline 3\end{array}\right)^3=1\] A beautiful solution, not by me :P – Daniel Liu · 1 year, 3 months ago

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Problem 6 :Let \(x > y > z\) be positive reals greater than one.

Prove that \[\large{ \sum_{cyc} x^{\frac{7}{3}} > \sum_{cyc} x^{2} y^{\frac{1}{3}}}\] – Harsh Shrivastava · 1 year, 4 months ago

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Solution 6: Thank You @Harsh Shrivastava,Since we are given \(x,y,z \in \mathbb{R^{+} }\) and \(x>y>z\).

We have two sets \(x^2 > y^2 > z^2\) and \(x^{\frac{1}{3}} > y^{\frac{1}{3}} > z^{\frac{1}{3}}\).

By rearrangement inequality we have

\[\Large{{ x }^{ 2 }{ x }^{ \frac { 1 }{ 3 } }+{ y }^{ 2 }{ y }^{ \frac { 1 }{ 3 } }+{ z }^{ 3 }{ z }^{ \frac { 1 }{ 3 } }>{ x }^{ 2 }{ y }^{ \frac { 1 }{ 3 } }+{ y }^{ 2 }{ z }^{ \frac { 1 }{ 3 } }+{ z }^{ 2 }{ x }^{ \frac { 1 }{ 3 } }\\ \therefore \quad \sum _{ \text{cyclic } }^{ }{ { x }^{ \frac { 7 }{ 3 } } } >\sum _{ \text{ cyclic } }^{ }{ { x }^{ 2 } } { y }^{ \frac { 1 }{ 3 } }}\]

Now my time to post questions!! Harsh By the way how did you thought of this? – Lakshya Sinha · 1 year, 4 months ago

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Problem 3Prove the following inequality if \(a,b,c\) are positive reals and \(a+b+c+abc=4\) ,

\[(1+\dfrac{a}{b} +ac)(1+\dfrac{b}{c} +ba)(1+\dfrac{c}{a} +cb) \geq 27\]

## Pakistan Round 1-2016

– Sualeh Asif · 1 year, 4 months agoLog in to reply

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\((4-a)(4-b)(4-c)\geq 27abc\)

\((64-16(a+b+c)+4(ab+bc+ca)-abc))\geq 27abc\)

\(ab+bc+ca\geq 3abc\)

Note that \(ab+bc+ca\geq 3(abc)^{\frac{2}{3}}\geq 3abc\)

This simplifies to \(1\geq abc\) which is true from our condition! – Sualeh Asif · 1 year, 4 months ago

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Problem 13Given that \(a,b,c\) are non-negative reals satisfying \(a+b+c=1\), prove that \[(1-a)(1-b)(1-c)\le \dfrac{7}{27}+abc\] – Daniel Liu · 1 year, 3 months ago

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Solution to Problem 13I expanded (once more) to get a solution.

Replace \((1-a), (1-b), (1-c)\) with \((b+c), (c+a), (a+b)\) in the in equation to get:

\[(a+b)(b+c)(c+a) \leq \dfrac {7}{27} + abc\]

Expanding and simplifying, we get

\[a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc \leq \dfrac {7}{27}\]

Removing the denominator, we get

\[27a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7\]

Homogenising the RHS, we get

\[27a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7(a+b+c)^3\]

\[27a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7a^3 + 7b^3 + 7c^3 + 21a^2b + 21a^2c + 21b^2a + 21b^2c + 21c^2a + 21c^2b + 42abc\]

\[6a^2b + 6a^2c + 6b^2a + 6b^2c + 6c^2a + 6c^2b \leq 7a^3+ 7b^3 + 7c^3 + 15abc\]

\[a^2(b + c) + b^2(a + c) + c^2(a + b) \leq \dfrac {7}{6} a^3 + \dfrac {7}{6} b^3 + \dfrac {7}{6} c^3 + \dfrac {5}{2} abc\]

Note that by Schur's, \(a^2(b + c) + b^2(a + c) + c^2(a + b) \leq a^3 + b^3 + c^3 + 3abc\). We will prove the sharper inequality:

\[a^3 + b^3 + c^3 + 3abc \leq \dfrac {7}{6} a^3 + \dfrac {7}{6} b^3 + \dfrac {7}{6} c^3 + \dfrac {5}{2} abc\]

\[6a^3 + 6b^3 + 6c^3 + 18abc \leq 7a^3 + 7b^3 + 7c^3 + 15abc\]

\[3abc \leq a^3 + b^3 + c^3\]

Which is obviously true by AM-GM. Thus, proven. – Sharky Kesa · 1 year, 3 months ago

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Extension: can you find the best constants for the RHS? – Daniel Liu · 1 year, 3 months ago

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Problem 25For the positive real numbers \(a, b, c\) prove that: \[\dfrac{1}{b(a+b)} + \dfrac{1}{c(b+c)} + \dfrac{1}{a(c+a)} \geq \dfrac{27}{2(a+b+c)^2}\] – Chris Galanis · 1 year, 3 months ago

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Solution to Problem 25This solution is quite long and extremely bashy. Sorry. :P

We have to prove

\[\displaystyle \sum_{\text{cyc}} \dfrac {1}{b(a+b)} \geq \dfrac {27}{2(a+b+c)^2}\]

Multiplying out the denominators and simplifying (here is the bash, but I have suppressed it), we get

\[\displaystyle \sum_{\text{cyc}} 2a^5 b + 6a^4 b^2 + 2a^2 b^4 + 6a^3 b^3 + 8a^4bc \geq \displaystyle \sum_{\text{cyc}} 7a^3 b^2 c + 9a^3 b c^2 + 8 a^2 b^2 c^2\]

By AM-GM, we have

\[\dfrac {a^5 b + a^4 b^2 + c^4 a^2 + a^2 b^4 + a^2 b^4 + a^3 b^3 + a^3 c^3}{7} \geq \sqrt[7]{a^{21}b^{14}c^7}\]

\[\dfrac {a^5 b + a^4 b^2 + c^4 a^2 + c^4 a^2 + c^4 a^2 + a^3 b^3 + a^3 b^3 + a^3 c^3 + a^3 c^3}{9} \geq \sqrt[9]{a^{27}b^{9}c^{18}}\]

\[\dfrac {8a^4bc + 8ab^4c + 8abc^4}{3} \geq \sqrt[3]{24a^6 b^6 c^6}\]

These inequations imply

\[a^5 b + a^4 b^2 + c^4 a^2 + 2a^2 b^4 + a^3 b^3 + a^3 c^3 \geq 7a^3 b^2 c\]

\[a^5 b + a^4 b^2 + 3c^4 a^2 + 2a^3 b^3 + 2a^3 c^3 \geq 9a^3 b c^2\]

\[8a^4bc + 8ab^4c + 8abc^4 \geq 24a^2 b^2 c^2\]

Adding the first two equations cyclicly, then adding the third equation, we get the statement we wished to prove. Thus proven!

PS: It took me hours to get the right grouping of terms for the AM-GM. – Sharky Kesa · 1 year, 3 months ago

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– Daniel Liu · 1 year, 3 months ago

You could have grouped them by triples of powers that majorize each other, and then applied AM-GM... Saves you all the time of finding your own grouping ;)Log in to reply

Problem 14Show that for non-negative real \(x\), \(y\) and \(z\),

\[2x^3 (x^3 + 8y^3) + 2y^3 (y^3 + 8z^3) + 2z^3 (z^3 + 8x^3) \geq 9x^4 (y^2 + z^2) + 9y^4 (z^2 + x^2) + 9z^4 (x^2 + y^2)\] – Sharky Kesa · 1 year, 3 months ago

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SOLUTION OF PROBLEM 14\[LHS - RHS = \displaystyle \sum_{cyc} (x-y)^4(x^2 + 4xy + y^2) \geq 0\] Hence proved – Chris Galanis · 1 year, 3 months agoLog in to reply

@Chris Galanis Can you please tell how you came up with that factorization? – Svatejas Shivakumar · 1 year, 3 months ago

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– Chris Galanis · 1 year, 3 months ago

Actually when I solve inequalities this way I expect \(LHS-RHS \geq 0\). Since \(x, y, z\) are non negative I expect all the negative terms to form a square. In particular I tried to find pattern like \(a^2 - 2ab +b^2\) (which is \((a-b)^2\)) or \(a^4 - 4a^3b + 6a^2b^2 -4ab^3 + b^4\) (which is \((a-b)^4\)). Just thatLog in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

Nice!Log in to reply

Problem 5Let \(x\), \(y\), \(z\) be positive reals such that \(xyz=1\). Show that

\[\dfrac {2}{(x+1)^2+y^2+1} + \dfrac {2}{(y+1)^2+z^2+1} + \dfrac {2}{(z+1)^2+x^2+1} \leq 1\] – Sharky Kesa · 1 year, 4 months ago

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Solution 5 :\((x-y)^{2} \geq 0 \)

\(\implies x^{2} + y^{2} \geq 2xy\)

Adding \(2x+2\) on both sides, \(x^{2} + y^{2} +2x + 2 \geq 2xy +2x +2\)

\(\implies \dfrac{1}{xy+x+1} \geq \dfrac{2}{x^{2} + y^{2} +2x + 2 }\)

\[\implies \dfrac{1}{xy+x+1} \geq \dfrac{2}{(x+1)^2+y^2+1 }\]

Similarly , \[\dfrac{1}{yz+y+1} \geq \dfrac{2}{(y+1)^2+z^2+1 }\] and \[\dfrac{1}{zx+z+1} \geq \dfrac{2}{(z+1)^2+x^2+1 }\]

Adding the three inequalities, we get \[\dfrac {2}{(x+1)^2+y^2+1} + \dfrac {2}{(y+1)^2+z^2+1} + \dfrac {2}{(z+1)^2+x^2+1} \leq \dfrac{1}{yz+y+1} + \dfrac{1}{xy+x+1}+ \dfrac{1}{xz+z+1} \]

Since \(\dfrac{1}{yz+y+1} + \dfrac{1}{xy+x+1}+ \dfrac{1}{xz+z+1} = 1 \), the result is evident. – Harsh Shrivastava · 1 year, 4 months ago

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@Chris Galanis, @Sualeh Asif, @Daniel Liu, @Sharky Kesa, @Harsh Shrivastava, @Svatejas Shivakumar. Thanks everyone fit making this such a beautiful contest, hope it went on! For season 2 I will be the organiser, comment please for it. – Lakshya Sinha · 1 year, 2 months ago

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– Harsh Shrivastava · 1 year, 2 months ago

Sure!Log in to reply

– Sualeh Asif · 1 year, 2 months ago

Waiting forit to begin...Log in to reply

– Svatejas Shivakumar · 1 year, 2 months ago

More than welcome :)Log in to reply

Problem 15Let \(x_1, x_2, x_3, \ldots, x_m >1\) ,where \(m \in \mathbb{N^+}\), be positive integers such that \(x_j < x_{j+1}\) \((1 \leq j < m)\).

Prove that \[\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} < 1\] – Chris Galanis · 1 year, 3 months ago

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We got \[x_i \geq i+1 \Leftrightarrow \frac{1}{x_i^3} \leq \frac{1}{(i+1)^3} < \frac{1}{(i+1)^2} < \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1} \Leftrightarrow \frac{1}{x_i^3} < \frac{1}{i} - \frac{1}{i+1}(*)\] Thus taking the sum for \(i = 1\) until \(i = m\): \[\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} \stackrel{(*)}{<} \sum_{i=1}^m \frac{1}{i} - \frac{1}{i+1} = 1 - \frac{1}{m+1} <1\] Since \(\displaystyle \sum_{i=1}^m \frac{1}{i} - \frac{1}{i+1}\) is a telescoping series. Hence proved – Chris Galanis · 1 year, 3 months ago

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– Daniel Liu · 1 year, 3 months ago

\[\sum_{i=1}^{m} \dfrac{1}{x_i^3} < \int_{i=1}^{\infty}\dfrac{1}{x^3}\text{ d}x = \dfrac{1}{2} < 1\]Log in to reply

PROBLEM 12If \(a,b,c,d\) are positive reals such that \(abcd=1\), prove that \(\dfrac{1}{(1+a)(1+a^{2})}+\dfrac{1}{(1+b)(1+b^{2})}+\dfrac{1}{(1+c)(1+c^{2})}+\dfrac{1}{(1+d)(1+d^{2})} \ge 1\) – Svatejas Shivakumar · 1 year, 3 months ago

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We want to prove \[\sum \dfrac{1}{1+a+a^2+a^3}\ge 1\] given \(abcd=1\). Let \(e^w=a, e^x=b\) etc. and let \(f(x)=\dfrac{1}{1+e^x+e^{2x}+e^{3x}}\) and the condition is \(w+x+y+z=0\) with the inequality \[f(w)+f(x)+f(y)+f(z) \ge 1\] By RCF Theorem since \(f(x)\) is convex for \(x\ge 0\) it suffices to prove the inequality for \(x=y=z\) and \(w=-3x\). So \[f(-3x)+3f(x)\ge 1\] Let \(g(x)=f(-3x)+3f(x)\) and taking the derivative gives us that \(x=0\) is the only local minima. Since \(\lim\limits_{x\to\infty}g(x)=1\) we need only to check \(x=0\). But \(g(0)=1\) so \(g(x) \ge 1\) and the inequality is proved. – Daniel Liu · 1 year, 3 months ago

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this link for the original solution. – Svatejas Shivakumar · 1 year, 3 months ago

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Problem 10Given that \(a_2, a_3, \ldots , a_n\) is a permutation of \(\{2, 3, \ldots , n\}\), prove that \[(2a_2-1)(3a_3-1)\cdots (na_n-1)\ge \dfrac{(n-1)!(n+1)!}{2}\] – Daniel Liu · 1 year, 3 months ago

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Solution 10 :Observe that\[\dfrac{(n-1)!(n+1)!}{2} = (n^{2}-1)((n-1)^{2} - 1) \cdots (3^{2}-1)(2^{2}-1)\]

So now the inequality becomes \[(2a_2-1)(3a_3-1)\cdots (na_n-1)\ge (n^{2}-1)((n-1)^{2} - 1) \cdots (3^{2}-1)(2^{2}-1)\]

This can be proven by applying Reverse Rearrangement Inequality on the sets \({X} : (a_{2}-1),(a_{3}-1) \cdots, (a_{n}-1)\) and \({Y} : (1 \times a_{2}), (2\times a_{3}),(3\times a_{4}) \cdots,( (n-1) \times a_{n})\).

\(X,Y\) will be similarly ordered when \(a_{k} = k\).

So applying Reverse Rearrangement Inequality (Random Sum Product > Similar ordered Product), \[\displaystyle\prod_{i=2}^n (X_{i}+Y_{i}) \geq \displaystyle\prod_{i=2}^n(i^{2} - 1)\] – Harsh Shrivastava · 1 year, 3 months ago

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Some of you might have seen this but still answer me.

Problem 7Let \(a, b, c\) be positive real numbers such that \( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\). Prove that

\[\Large{\frac { 1 }{ { \left( 2a+b+c \right) }^{ 2 } } +\frac { 1 }{ { \left( 2b+c+a \right) }^{ 2 } } +\frac { 1 }{ { \left( 2c+a+b \right) }^{ 2 } } \le \frac { 3 }{ 16 } }\] – Lakshya Sinha · 1 year, 4 months ago

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Solution to Problem 7By AM-GM \(\dfrac{1}{(2a+b+c)^2} \le \dfrac{1}{4(a+b)(a+c)}\) and thus it remains to prove \[\sum_{cyc} \dfrac{1}{(a+b)(a+c)}\le \dfrac{3}{4}\] Clearing denominators, it remains to prove \(8(a+b+c)\le 3(a+b)(b+c)(c+a)\) and after homogenizing this becomes \[8abc(a+b+c)^2 \le 3(ab+bc+ca)(a+b)(b+c)(c+a)\] which after expansion and simplification is \[3\sum_{sym}a^3b^2 \ge \sum_{sym}a^3bc+2\sum_{sym}a^2b^2c\] which is true by Muirhead. – Daniel Liu · 1 year, 4 months ago

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PROBLEM 2 :Let \(a,b\) and \(c\) be non-negative reals such that \(a+b+c = 1\).

Show that \[1 + 6abc \geq \dfrac{1}{4} + 3(a+b)(b+c)(c+a)\] – Harsh Shrivastava · 1 year, 4 months ago

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– Daniel Liu · 1 year, 4 months ago

Homogenizing and expanding gives \[a^3+b^3+c^3+6abc \ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\] which is just Schur.Log in to reply

Solution Problem 2:\(1+6abc \geq \dfrac{1}{4} +3(1-a)(1-b)(1-c)\)

\(1+6abc \geq \dfrac{1}{4} +3(1-(a+b+c)+\sum_{cyc} ab -abc)\)

\(1+6abc \geq \dfrac{1}{4} +3(\sum_{cyc} ab -abc)\)

\(\dfrac{3}{4}+9abc \geq 3 \times \sum_{cyc} ab \)

\(\dfrac{1}{4}+3abc - \sum_{cyc} ab \geq 0\)

\(1+ 12abc \geq 4 \sum_{cyc} ab\)

\((a+b+c)^3+12abc \geq \sum_{cyc} ab\)

\(a^3+b^3+c^3+6abc \geq \sum_{sym} a^2b\)

This is schurs inequality. – Sualeh Asif · 1 year, 4 months ago

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– Daniel Liu · 1 year, 4 months ago

How did you use \(ab+bc+ca \le \dfrac{1}{3}\) on \(1+12abc\ge 4(ab+bc+ca)\) to get \(abc\ge 0\)?Log in to reply

@Sharky Kesa The contest ended? Wish it could have continued for some for time. Anyway thanks to everyone who participated in this contest, learnt a lot and had a lot of fun. – Svatejas Shivakumar · 1 year, 2 months ago

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Problem 27Given positive reals \(a,b,c\), prove that \[3\sum_{cyc}\dfrac{a+b}{c}\ge 10+8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}\] – Daniel Liu · 1 year, 3 months ago

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(1)\((a+b)(b+c)(c+a) \ge \dfrac{8}{9}(a+b+c)(ab+bc+ca)\) which is true since by expanding and simplifying we get \(\displaystyle \sum_{sym} a^2b \ge 6abc\) which is rearrangement inequality.(2)\(\Big(\displaystyle \sum_{sym} a^2b\Big)+2abc = (a+b)(b+c)(c+a) \) and \((a+b+c)^2 = a^2+b^2+c^2 +2(ab+bc+ca)\)\[3 \displaystyle \sum_{cyc} \dfrac{a+b}{c} \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}\] Adding \(6\) to both sides and using

(2)we get:\[3\dfrac{(a+b)(b+c)(c+a)}{abc} \ge 8\dfrac{(a+b+c)^2}{ab+bc+ca}\] Thus by

(1)it suffices to prove that:\[(ab+bc+ca)^2\ge 3abc(a+b+c)\] Which is true since by expanding and simplifying both sides we get rearrangement inequality. Hence proved! – Chris Galanis · 1 year, 3 months ago

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Problem 26Given positive reals \(x\), \(y\), \(z\) greater than or equal to 1 such that

\[\dfrac {1}{x} + \dfrac {1}{y} + \dfrac {1}{z} = 2\]

Prove

\[\sqrt{x+y+z} \geq \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}\] – Sharky Kesa · 1 year, 3 months ago

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The problem is equivalent to \[(x+y+z)\left(3-\dfrac{1}{x}-\dfrac{1}{y}-\dfrac{1}{z}\right)\ge \left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2\]

Confirm that \(1-\dfrac{1}{x}\ge 0\implies x\ge 1\); thus, by C-S we are done. – Daniel Liu · 1 year, 3 months ago

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Sorry for the delay guys, and the inconvenience caused by me.

Problem 24:Let a,b,c be positive reals.

Prove that \[\displaystyle \prod_{cyc} (1 + \dfrac{a}{b}) \geq 2(1 + \dfrac{a+b+c}{ (abc)^{\frac{1}{3}}})\] – Harsh Shrivastava · 1 year, 3 months ago

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Substitute \(a = x^{3}, b= y^{3} and c = z^{3}\), we get \[\large \sum_{cyc} (\frac{x^{3}}{y^{3}} + \frac{x^{3}}{z^{3}}) \geq 2\dfrac{(x^{3} + y^{3}+ z^{3})}{xyz} = \sum_{cyc} (\dfrac{x^{2}}{yz} + \dfrac{x^{2}}{yz})\]

Now using re-arrangement inequality, we can get our desired result. – Harsh Shrivastava · 1 year, 3 months ago

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By applying Holder's Inequality to LHS it suffices to prove that \[\Big((1+(\dfrac{a}{b} \dfrac{b}{c} \dfrac{c}{a})^{1/3}\Big)^3 \geq 2(1+\dfrac{a+b+c}{(abc)^{1/3}}) \\ \Rightarrow 4 \geq 1+\dfrac{a+b+c}{(abc)^{1/3}} \\ \Rightarrow (abc)^{1/3} \geq \dfrac{a+b+c}{3}\] But by AM-GM this is true only for \(a=b=c=1\)

Edit: If the problem had the direction of the inequality flipped then we would have to prove that

\[\begin{equation} \begin{split} (1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a}) & \leq 2(1+\dfrac{a+b+c}{(abc)^{1/3}}) \\ & \stackrel{AM-GM}{\leq}2(1+\dfrac{a+b+c}{{a+b+c}/_3}) = 8\end{split} \end{equation}\]

Then it suffices to prove that \[\begin{equation} \begin{split} & (a+b)(b+c)(c+a) \leq 8abc \\ & \stackrel{AM-GM}{\leq} \Big(\dfrac{2}{3}(a+b+c)\Big)^3 \leq 8abc \\ & \Rightarrow a^3 + b^3 +c^3 +3a^2b+3a^2c+3ab^2+3ac^2+3b^2c+3bc^2 \leq 21abc \end{split} \end{equation}\] But again that would be true by rearrangement inequality if the direction of the inequality was flipped... – Chris Galanis · 1 year, 3 months ago

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– Daniel Liu · 1 year, 3 months ago

The problem is that every time you apply an inequality, the inequality weakens a bit. If you apply inequalities in the incorrect fashion or too much, then the original inequality will weaken to the point that it is false.Log in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

Next problem please!Log in to reply

– Harsh Shrivastava · 1 year, 3 months ago

I request someone to post a problem on my behalf. Thanks!Log in to reply

@Chris Galanis I hope you post your problem soon! – Sualeh Asif · 1 year, 3 months ago

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– Chris Galanis · 1 year, 3 months ago

This was not a solution... I wait for Harsh or someone else to post the solutionLog in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

I believe that it isn't a solution. The sign must be interchanged. Even I got that after applying A.M-G.M. You pretty much cannot use anything I guess. Such a strange inequality.Log in to reply

We can factor it to \[\sum_{cyc} (a+b)(ab-1)\geq 0\] – Sualeh Asif · 1 year, 3 months ago

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– Harsh Shrivastava · 1 year, 3 months ago

Hmm sorry guyz I won't be able to post solution now because I am a bit busy. Will post a solution by tmmrw evening(my time zone).Log in to reply

Problem 23Find the greatest positive real number \(k\), for which is true that \[\dfrac{xy}{\sqrt{(x^2+y^2)(3x^2+y^2)}} \leq \dfrac{1}{k}\] for all positive real numbers \(x, y\). – Chris Galanis · 1 year, 3 months ago

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Problem 22Here is a very nice problem I came across.

Given that \(a,b,c\) are non-negative reals, prove that \[(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\] – Daniel Liu · 1 year, 3 months ago

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Solution of Problem 22By Cauchy we got: \((a^2+2)(1+\frac{(b+c)^2}{2}) \geq (a+b+c)^2\)

Then it suffices to prove that \((b^2+2)(c^2+2) \geq 3(1+\frac{(b+c)^2}{2})\)

By rearranging and simplifying we get: \[(b^2-2bc+c^2) + (2b^2c^2-4bc+1) \geq 0 \\ (b-c)^2 + 2(bc-1)^2 \geq 0\] Which is true. Equality holds iff \(a=b=c=1\). – Chris Galanis · 1 year, 3 months ago

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– Daniel Liu · 1 year, 3 months ago

It's false ;)Log in to reply

– Chris Galanis · 1 year, 3 months ago

Can you show that please?Log in to reply

EDIT: I tried to solve the problem using your idea and it appears that you have only just typoed. Splendid solution! – Daniel Liu · 1 year, 3 months ago

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– Chris Galanis · 1 year, 3 months ago

Sorry, I fixed that !Log in to reply

Problem 16For positive reals \(a,b,c\), prove \[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1. \]

Source: ISL – Daniel Liu · 1 year, 3 months ago

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Solution to Problem 16I used Jensen's Inequality to solve this problem.

Apply Jensen's on the convex function

\[f(x) = \dfrac {1}{\sqrt{x}}\]

and weights \(a\), \(b\), \(c\) in the following way.

\[\dfrac {a \text{ } f(a^2 + 8bc) + b \text{ } f(b^2 + 8ca) + c \text{ } f(c^2 + 8ab)}{a+b+c} \geq f \left (\dfrac {a(a^2 + 8bc) + b(b^2 + 8ca) + c(c^2 + 8ab)}{a+b+c} \right )\]

This may be rewritten as

\[\dfrac{a}{\sqrt{a^2 + 8bc}} + \dfrac{b}{\sqrt{b^2 + 8ca}} + \dfrac{c}{\sqrt{c^2 + 8ab}} \geq \dfrac {(a+b+c)^{\frac {3}{2}}}{\sqrt {a^3 + b^3 + c^3 + 24abc}}.\]

Thus, it suffices to prove

\[(a+b+c)^3 \geq a^3 + b^3 + c^3 + 24abc\]

But we can expand this and simplify to get

\[a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc \geq 8abc\]

\[(a+b)(b+c)(c+a) \geq 8abc\]

Which is true by AM-GM. Thus proven. – Sharky Kesa · 1 year, 3 months ago

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Problem 11 :If \(a,b,c\) are positive real numbers that satisfy \(a+b+c=1\), prove that \[\sqrt{a+bc}+ \sqrt{b+ac} + \sqrt{c+ba} \leq 2 \] – Harsh Shrivastava · 1 year, 3 months ago

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SOLUTION OF PROBLEM 11\((a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=1\) or \(ab+bc+ca \le \dfrac{1}{3}\) \(\color{Red}{\text{Since}} \quad \color{Red}{{a^{2}+b^{2}+c^{2} \ge ab+bc+ca}}\)

Now, by Cauchy-Schwarz inequality \(\sqrt{\left((\sqrt{a+bc})^{2}+(\sqrt{b+ac})^{2}+(\sqrt{c+ba})^{2} \right)(1+1+1)} \ge \sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba}\) or \(\sqrt{3+3(ab+bc+ca)} \ge \sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba}\)

Now \(\sqrt{3+3(ab+bc+ca)} \le 2\) since \(ab+bc+ca \le \dfrac{1}{3}\). Hence \(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba} \le 2\).

Equality holds when \(a=b=c=\dfrac{1}{3}\). Hence proved. – Svatejas Shivakumar · 1 year, 3 months ago

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– Chris Galanis · 1 year, 3 months ago

Can you please elaborate more on how you applied Cauchy-Schwarz here?Log in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

It's a just applying Cauchy-Schwarz with \(a+bc,b+ac,c+ba\) and (1,1,1) then, taking square root on both sides.Log in to reply

Problem 28For the real numbers \(a, b, c, d\) prove that: \[-1\le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1\] – Chris Galanis · 1 year, 3 months ago

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Now it suffices to consider the case when one of \(a,b,c,d\) are negative. If one of them is negative. WLOG assume \(a\) then let \(A=-a\). Thus \[(bd-Ac)^2\leq (bd+ac)^2 \leq (a^2+b^2)(c^2+d^2)\] equality holds when \(A=a=0\)

If any two of them are negative we have two cases .

WLOG assume \(a,b\). Then let \(A=-a\), \(B=-b\), \(\therefore (Ac+Bd)^2=(ac+bd)^2\) Otherwise assume both of \(a,c\) are negative.Then there multiple is positive and we are done.

The case when 3 are negative is equivalent to the case when one is negative. – Sualeh Asif · 1 year, 3 months ago

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– Chris Galanis · 1 year, 3 months ago

Yet that is not a proof since \(a, b, c, d\) may not be positiveLog in to reply

I tend to keep posting my progress because the mobile app gets stuck – Sualeh Asif · 1 year, 3 months ago

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Problem 20Let \(a\), \(b\) and \(c\) be positive reals such that \(abc=8\). Prove that

\[\dfrac {a^2}{\sqrt{(1+a^3)(1+b^3)}} + \dfrac {b^2}{\sqrt{(1+b^3)(1+c^3)}} + \dfrac {c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \dfrac {4}{3} .\] – Sharky Kesa · 1 year, 3 months ago

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Solution of Problem 20Note that \(\sqrt{1+k^3} = \sqrt{(1+k)(1-k+k^2)} \stackrel{AM-GM}{\leq} \frac{k^2-k+1+k+1}{2} = \frac{k^2+2}{2} \Leftrightarrow \frac{2}{k^2+2} \leq \frac{1}{\sqrt{(1+k^3)}} (*)\) Hence: \[\begin{equation} \begin{split} \displaystyle \sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} & \geq \frac{4}{3} \\ \stackrel{(*)}{\Rightarrow} \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} & \geq \frac{4}{3} \\ \Rightarrow 3 \sum_{cyc} a^2(c^2+2) & \geq (a^2+2)(b^2+2)(c^2+2) \\ \Rightarrow \sum_{cyc} 2a^2+ \sum_{cyc} a^2b^2 & \geq 72 \end{split} \end{equation}\] Which is true by applying AM-GM in both sums separately. Hence proved. – Chris Galanis · 1 year, 3 months ago

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Problem 19 :Let \(x,y,z\) be positive real numbers.Prove that \[ \displaystyle \sum_{cyc} \dfrac{1}{x^{3} + y^{3} + xyz} \leq \dfrac{1}{xyz}\] – Harsh Shrivastava · 1 year, 3 months ago

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Solution to Problem 19Since this inequality is homogenous, WLOG \(xyz = 1\). We will prove that

\[\dfrac {1}{x^3+y^3+1} \leq \dfrac {z}{x+y+z}\]

\[(x^3+y^3)z+z \geq x+y+z\]

Note that \(x^3+y^3 = (x+y)(x^2-xy+y^2)\).

\[(x+y)(x^2-xy+y^2)z \geq x+y\]

\[(x^2 + y^2 - xy)z \geq 1\]

By AM-GM, \(x^2 + y^2 \geq 2xy\). Substituting, we get

\[(2xy-xy)z \geq 1\]

\[xyz \geq 1\]

Which is obviously true by our assumption. Thus, proven. The inequality in the question can be achieved by adding cyclicly.

We have already proven (with WLOG \(xyz=1\) because of the equation's homogeneity)

\[\dfrac {1}{x^3 + y^3 + 1} \leq \dfrac {z}{x+y+z}\]

This means that

\[\dfrac {1}{x^3 + y^3 + 1} + \dfrac {1}{y^3 + z^3 + 1} + \dfrac {1}{z^3 + x^3 + 1} \leq \dfrac {x+y+z}{x+y+z}\]

\[\displaystyle \sum_{\text{cyc}} \dfrac {1}{x^3 + y^3 + 1} \leq 1\]

Note that since \(xyz=1\), \(\dfrac {1}{xyz} = 1\). Thus,

\[\displaystyle \sum_{\text{cyc}} \dfrac {1}{x^3 + y^3 + xyz} \leq \dfrac {1}{xyz}\] – Sharky Kesa · 1 year, 3 months ago

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PROBLEM 18For positive real numbers \(a,b,c\) with \(a+b+c=abc\), prove that \(\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}\). – Svatejas Shivakumar · 1 year, 3 months ago

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Solution Of Problem 18 :Substitute a= tanx , b= tany , c= tanz such that \( x+y+z= \pi\)

Since a,b,c are positive reals, x,y,z are all less than \( \frac{\pi}{2}\)

Now we want to prove that \( \cos x + \cos y + \cos z \leq 1.5\)

By Jensen's Inequality,

\[\cos \left(\dfrac{x+y+z}{3}\right)\ge \dfrac{\cos x+\cos y+\cos z}{3}\]

\( \implies \cos x + \cos y + \cos z \leq \dfrac{3}{2}\)

Note that \(f(p) = \cos p\) is concave down down for all \( 0 \leq p \leq \frac{\pi}{2} \).This can be checked by differentiating f(p) twice. – Harsh Shrivastava · 1 year, 3 months ago

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Problem 17For positive reals \(x\), \(y\) and \(z\) such that \(xyz=1\), prove that

\[\dfrac {1}{x^3(y+z)} + \dfrac {1}{y^3(z+x)} + \dfrac {1}{z^3 (x+y)} \geq \dfrac {3}{2}\] – Sharky Kesa · 1 year, 3 months ago

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SOLUTION OF PROBLEM 17Dividing the numerator and denoninator by \(x^2,y^2\) and \(z^2\) respectively and applying Titu's Lemma we get \(\dfrac {1}{x^3(y+z)} + \dfrac{1}{y^3(z+x)}+\dfrac{1}{z^3 (x+y)} \geq \dfrac{(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}{2(xy + yz+ zx)}=\dfrac{(xy+yz+zx)^2}{2(xy+yz+zx)}=\dfrac{xy+yz+zx}{2}\)

Now, by A.M-G.M inequality \(\dfrac{xy+yz+zx}{2} \geq 3 \cdot \dfrac{\sqrt[3]{(xyz)^2}}{2}=\dfrac{3}{2}\).

Equality holds if and only if \(x=y=z=1\). Hence proved. – Svatejas Shivakumar · 1 year, 3 months ago

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Problem 9Let \(x\), \(y\), \(z\) be reals such that \(xyz = -1\). Show that

\[x(x^3+3) + y(y^3+3) + z(z^3+3) \geq \dfrac {x^2}{y} + \dfrac {x^2}{z} + \dfrac {y^2}{x} + \dfrac {y^2}{z} + \dfrac {z^2}{x} + \dfrac {z^2}{y}\] – Sharky Kesa · 1 year, 3 months ago

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Homogenizing and expanding reduces it to \[\sum_{sym}x^4+2\sum_{sym}x^3y\ge 3\sum_{sym} x^2yz\] which factors as \[\dfrac{1}{2}\sum_{sym}(x^2-y^2)^2+\sum_{sym}xy(x-y)^2+\dfrac{3}{2}\sum_{sym}x^2(y-z)^2\ge 0\] which is true. – Daniel Liu · 1 year, 3 months ago

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Now I think what is left to prove is that: \[\frac{(x+y+z)^4}{27} + 3(x+y+z) \geq \frac{x^2}{y} +\frac{x^2}{z} +\frac{y^2}{x} +\frac{y^2}{z} +\frac{z^2}{x} +\frac{z^2}{y} \] But I don't know how to go on. Am I correct this far? – Chris Galanis · 1 year, 3 months ago

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– Lakshya Sinha · 1 year, 3 months ago

Yeah I think soLog in to reply

PROBLEM 4For positive real numbers \(a,b,c\) prove that \(\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a} \le 1\) – Svatejas Shivakumar · 1 year, 4 months ago

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\[\dfrac {a}{2a+b} + \dfrac {b}{2b+c} + \dfrac {c}{2c+a} \leq 1\]

\[a(2b+c)(2c+a) + b(2a+b)(2c+a)+c(2a+b)(2b+c) \leq (2a+b)(2b+c)(2c+a)\]

\[12abc+4a^2b+4b^2c+4c^2a+ab^2+bc^2+ca^2 \leq 9abc+4a^2b+4b^2c+4c^2a+2ab^2+2bc^2+2ca^2\]

\[3abc \leq ab^2+bc^2+ca^2\]

The last statement is obviously true by AM-GM. Thus, proven. – Sharky Kesa · 1 year, 4 months ago

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Problem 21If for the real numbers \(a, b, c\) where \(bc \neq 0\) is true that \(\frac{1-c^2}{bc} \geq 0\), prove that: \[10(a^2+b^2+c^2-bc^3) \geq 2ab + 5ac\] – Chris Galanis · 1 year, 3 months ago

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If \(|c| \le 1\) and \(\text{sgn}(b) = \text{sgn}(c)\), then the inequality is \[10(a^2+b^2+c^2-bc^3)\ge 10(a^2+b^2+c^2-bc)\ge 2ab+5ac\]\[\iff (a-b)^2+\dfrac{5}{2}(a-c)^2+5(b-c)^2+\dfrac{13}{2}a^2+4b^2+\dfrac{5}{2}c^2\ge 0\] done.

Not a very strong inequality; no equality case in fact. – Daniel Liu · 1 year, 3 months ago

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– Svatejas Shivakumar · 1 year, 3 months ago

Can this problem be done without calculus? I have no idea how to proceed with this problem.Log in to reply

– Lakshya Sinha · 1 year, 3 months ago

Me and Sharky were discussing about it and it showed this can be nearly solved using \(x^2 \ge 0, x\in \mathbb{R}\).Log in to reply