# Brilliant Inequality Contest - Season 1

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.

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

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

3. 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.)

4. 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.

5. 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.

6. The scope of the problems is Olympiad-style inequalities.

7. 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).

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

9. 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.

10. 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.

SOLUTION OF PROBLEM (insert problem number here)

PROBLEM (insert problem number here)

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

PROBLEM 1

Let $$x$$, $$y$$ 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.

Note by Sharky Kesa
3 years ago

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Solution 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,z are sides 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,z are not sides 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.

- 3 years ago

Can you write the equality cases for Case 2?

- 3 years ago

I don't think equality exists . . .

- 3 years ago

Problem 8

For positive reals $$a,b,c$$, prove that $\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\le \dfrac{1}{abc}$

- 3 years ago

Solution to Problem 8

I 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.

- 3 years ago

Consider this:

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

- 3 years ago

Problem 3
Prove 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

- 3 years ago

Problem 6 :

Let $$x > y > z$$ be positive reals greater than one.

Prove that $\large{ \sum_{cyc} x^{\frac{7}{3}} > \sum_{cyc} x^{2} y^{\frac{1}{3}}}$

- 3 years ago

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?

- 3 years ago

Problem 13

Given that $$a,b,c$$ are non-negative reals satisfying $$a+b+c=1$$, prove that $(1-a)(1-b)(1-c)\le \dfrac{7}{27}+abc$

- 3 years ago

Solution to Problem 13

I 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.

- 3 years ago

Note that this problem is actually equivalent to IMO 1984 #1.

Extension: can you find the best constants for the RHS?

- 3 years ago

Problem 5

Let $$x$$, $$y$$, $$z$$ be positive reals such that $$xyz=1$$. Show that

$\dfrac {2}{(x+1)^2+y^2+1} + \dfrac {2}{(y+1)^2+z^2+1} + \dfrac {2}{(z+1)^2+x^2+1} \leq 1$

- 3 years ago

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.

- 3 years ago

Problem 14

Show that for non-negative real $$x$$, $$y$$ and $$z$$,

$2x^3 (x^3 + 8y^3) + 2y^3 (y^3 + 8z^3) + 2z^3 (z^3 + 8x^3) \geq 9x^4 (y^2 + z^2) + 9y^4 (z^2 + x^2) + 9z^4 (x^2 + y^2)$

- 3 years ago

SOLUTION OF PROBLEM 14 $LHS - RHS = \displaystyle \sum_{cyc} (x-y)^4(x^2 + 4xy + y^2) \geq 0$ Hence proved

- 3 years ago

@Chris Galanis Can you please tell how you came up with that factorization?

- 3 years 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 that

- 3 years ago

Nice!

- 3 years ago

Problem 25

For the positive real numbers $$a, b, c$$ prove that: $\dfrac{1}{b(a+b)} + \dfrac{1}{c(b+c)} + \dfrac{1}{a(c+a)} \geq \dfrac{27}{2(a+b+c)^2}$

- 3 years ago

Solution to Problem 25

This 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.

- 3 years 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 ;)

- 3 years ago

PROBLEM 2 :

Let $$a,b$$ and $$c$$ be non-negative reals such that $$a+b+c = 1$$.

Show that $1 + 6abc \geq \dfrac{1}{4} + 3(a+b)(b+c)(c+a)$

- 3 years 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.

- 3 years ago

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.

- 3 years ago

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

- 3 years ago

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

Problem 7

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

$\Large{\frac { 1 }{ { \left( 2a+b+c \right) }^{ 2 } } +\frac { 1 }{ { \left( 2b+c+a \right) }^{ 2 } } +\frac { 1 }{ { \left( 2c+a+b \right) }^{ 2 } } \le \frac { 3 }{ 16 } }$

- 3 years ago

Solution to Problem 7

By 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.

- 3 years ago

Problem 10

Given that $$a_2, a_3, \ldots , a_n$$ is a permutation of $$\{2, 3, \ldots , n\}$$, prove that $(2a_2-1)(3a_3-1)\cdots (na_n-1)\ge \dfrac{(n-1)!(n+1)!}{2}$

- 3 years ago

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), $\displaystyle\prod_{i=2}^n (X_{i}+Y_{i}) \geq \displaystyle\prod_{i=2}^n(i^{2} - 1)$

- 3 years ago

PROBLEM 12

If $$a,b,c,d$$ are positive reals such that $$abcd=1$$, prove that $$\dfrac{1}{(1+a)(1+a^{2})}+\dfrac{1}{(1+b)(1+b^{2})}+\dfrac{1}{(1+c)(1+c^{2})}+\dfrac{1}{(1+d)(1+d^{2})} \ge 1$$

- 3 years ago

Problem 15

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

Prove that $\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} < 1$

- 3 years ago

Another interesting solution.

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

- 3 years ago

$\sum_{i=1}^{m} \dfrac{1}{x_i^3} < \int_{i=1}^{\infty}\dfrac{1}{x^3}\text{ d}x = \dfrac{1}{2} < 1$

- 3 years ago

@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.

- 2 years, 11 months ago

Sure!

- 2 years, 11 months ago

More than welcome :)

- 2 years, 11 months ago

Waiting forit to begin...

- 2 years, 11 months ago

Problem 11 :

If $$a,b,c$$ are positive real numbers that satisfy $$a+b+c=1$$, prove that $\sqrt{a+bc}+ \sqrt{b+ac} + \sqrt{c+ba} \leq 2$

- 3 years ago

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.

- 3 years ago

Can you please elaborate more on how you applied Cauchy-Schwarz here?

- 3 years 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.

- 3 years ago

Problem 16

For 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

- 3 years ago

Solution to Problem 16

I 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.

- 3 years ago

Problem 22

Here is a very nice problem I came across.

Given that $$a,b,c$$ are non-negative reals, prove that $(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$

- 3 years ago

Solution of Problem 22

By 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 $$a=b=c=1$$.

- 3 years ago

Can you please elaborate on your first step? I don't seem to see the Cauchy. Did you perhaps mean $$\dfrac{(b+c)^2}{2}$$ instead of $$\dfrac{b+c}{2}$$?

EDIT: I tried to solve the problem using your idea and it appears that you have only just typoed. Splendid solution!

- 3 years ago

Sorry, I fixed that !

- 3 years ago

Problem 23

Find 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 $$x, y$$.

- 3 years ago

- 3 years ago

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

Problem 24:

Let a,b,c be positive reals.

Prove that $\displaystyle \prod_{cyc} (1 + \dfrac{a}{b}) \geq 2(1 + \dfrac{a+b+c}{ (abc)^{\frac{1}{3}}})$

- 3 years ago

Open up the brackets and simplify a bit to get $\large \sum_{cyc} (\frac{a}{b} + \frac{a}{c}) \geq 2\dfrac{(a+b+c)}{(abc)^{\frac{1}{3}}}$

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.

- 3 years ago

I think this is equivalent to $$(1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a}) \geq 2(1+\dfrac{a+b+c}{(abc)^{1/3}})$$

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

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

Then it suffices to prove that $$$\begin{split} & (a+b)(b+c)(c+a) \leq 8abc \\ & \stackrel{AM-GM}{\leq} \Big(\dfrac{2}{3}(a+b+c)\Big)^3 \leq 8abc \\ & \Rightarrow a^3 + b^3 +c^3 +3a^2b+3a^2c+3ab^2+3ac^2+3b^2c+3bc^2 \leq 21abc \end{split}$$$ But again that would be true by rearrangement inequality if the direction of the inequality was flipped...

- 3 years 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.

- 3 years ago

@Chris Galanis I hope you post your problem soon!

- 3 years 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.

- 3 years ago

Since the inequality is homogenous, we assume $$abc=1$$.
We can factor it to $\sum_{cyc} (a+b)(ab-1)\geq 0$

- 3 years ago

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

- 3 years ago

- 3 years ago

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

- 3 years 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).

- 3 years ago

Problem 26

Given positive reals $$x$$, $$y$$, $$z$$ greater than or equal to 1 such that

$\dfrac {1}{x} + \dfrac {1}{y} + \dfrac {1}{z} = 2$

Prove

$\sqrt{x+y+z} \geq \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}$

- 3 years ago

Another solution:

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.

- 3 years ago

Problem 27

Given positive reals $$a,b,c$$, prove that $3\sum_{cyc}\dfrac{a+b}{c}\ge 10+8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$

- 3 years ago

(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!

- 2 years, 12 months ago

@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.

- 2 years, 11 months ago

PROBLEM 4

For positive real numbers $$a,b,c$$ prove that $$\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a} \le 1$$

- 3 years ago

I have a long way of doing this (Long because I expanded).

$\dfrac {a}{2a+b} + \dfrac {b}{2b+c} + \dfrac {c}{2c+a} \leq 1$

$a(2b+c)(2c+a) + b(2a+b)(2c+a)+c(2a+b)(2b+c) \leq (2a+b)(2b+c)(2c+a)$

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

$3abc \leq ab^2+bc^2+ca^2$

The last statement is obviously true by AM-GM. Thus, proven.

- 3 years ago

Problem 9

Let $$x$$, $$y$$, $$z$$ be reals such that $$xyz = -1$$. Show that

$x(x^3+3) + y(y^3+3) + z(z^3+3) \geq \dfrac {x^2}{y} + \dfrac {x^2}{z} + \dfrac {y^2}{x} + \dfrac {y^2}{z} + \dfrac {z^2}{x} + \dfrac {z^2}{y}$

- 3 years ago

What a strange inequality.

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.

- 3 years ago

Applying Jensen's inequality in LHS we get $\frac{f(x) + f(y) + f(z)}{3} \geq f\bigg(\frac{x+y+z}{3}\bigg) \\ \Rightarrow f(x) + f(y) + f(z) \geq 3\cdot \Bigg(\bigg(\frac{x+y+z}{3}\bigg)^4 + 3\cdot \bigg(\frac{x+y+z}{3}\bigg)\Bigg) \\ \Rightarrow \displaystyle \sum_{cyc} x(x^3+3) \geq \frac{(x+y+z)^4}{27} + 3(x+y+z)$ for $$f(x) = x(x^3 +3) = x^4 + 3x$$ which is convex as its graph is U-shaped

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?

- 3 years ago

Yeah I think so

- 3 years ago

Problem 17

For positive reals $$x$$, $$y$$ and $$z$$ such that $$xyz=1$$, prove that

$\dfrac {1}{x^3(y+z)} + \dfrac {1}{y^3(z+x)} + \dfrac {1}{z^3 (x+y)} \geq \dfrac {3}{2}$

- 3 years ago

This is a very famous inequality that appeared in IMO in the late 90s I believe. My approach goes as follows.

SOLUTION OF PROBLEM 17

Dividing 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.

- 3 years ago

PROBLEM 18

For positive real numbers $$a,b,c$$ with $$a+b+c=abc$$, prove that $$\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}$$.

- 3 years ago

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.

- 3 years ago

Problem 19 :

Let $$x,y,z$$ be positive real numbers.Prove that $\displaystyle \sum_{cyc} \dfrac{1}{x^{3} + y^{3} + xyz} \leq \dfrac{1}{xyz}$

- 3 years ago

Solution to Problem 19

Since 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,

$\displaystyle \sum_{\text{cyc}} \dfrac {1}{x^3 + y^3 + xyz} \leq \dfrac {1}{xyz}$

- 3 years ago

Problem 20

Let $$a$$, $$b$$ and $$c$$ be positive reals such that $$abc=8$$. Prove that

$\dfrac {a^2}{\sqrt{(1+a^3)(1+b^3)}} + \dfrac {b^2}{\sqrt{(1+b^3)(1+c^3)}} + \dfrac {c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \dfrac {4}{3} .$

- 3 years ago

Solution of Problem 20

Note 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: $$$\begin{split} \displaystyle \sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} & \geq \frac{4}{3} \\ \stackrel{(*)}{\Rightarrow} \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} & \geq \frac{4}{3} \\ \Rightarrow 3 \sum_{cyc} a^2(c^2+2) & \geq (a^2+2)(b^2+2)(c^2+2) \\ \Rightarrow \sum_{cyc} 2a^2+ \sum_{cyc} a^2b^2 & \geq 72 \end{split}$$$ Which is true by applying AM-GM in both sums separately. Hence proved.

- 3 years ago

Problem 28

For the real numbers $$a, b, c, d$$ prove that: $-1\le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$

- 2 years, 12 months ago

For the right hand side note that for positive real numbers(or all negative) $$(a^2 +b^2)(c^2+d^2)\geq (ac+bd)^2$$ By C.S .

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.

- 2 years, 12 months ago

Yet that is not a proof since $$a, b, c, d$$ may not be positive

- 2 years, 12 months ago

Of course completed the solution.

I tend to keep posting my progress because the mobile app gets stuck

- 2 years, 12 months ago

Problem 21

If for the real numbers $$a, b, c$$ where $$bc \neq 0$$ is true that $$\frac{1-c^2}{bc} \geq 0$$, prove that: $10(a^2+b^2+c^2-bc^3) \geq 2ab + 5ac$

- 3 years ago

If $$|c| \ge 1$$ 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)\ge 2ab+5ac$$\iff (a-b)^2+\dfrac{5}{2}(a-c)^2+\dfrac{13}{2}a^2+9b^2+\dfrac{15}{2}c^2\ge 0$ solved.

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.

- 3 years ago

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

- 3 years ago

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

- 3 years ago