Brilliant Inequality Contest - Season 1 - Sharky Kesa

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 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.

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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 $\color{#3D99F6}{x=a+b,y=b+c,z=a+c}$ ; $\text{(a,b,c>0)}$ ,so our inequality becomes:

$\color{#69047E}{\sum_{cyc}2(a+b)^2(a+b+2c) \geq \sum_{cyc} (a+b)^3 +9(a+b)(b+c)(a+c)}$

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

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

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

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

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

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

$\color{#D61F06}{\implies \sum_{cyc} a^{3} + 3abc \geq a^{2} b + b^{2} a + a^{2} c + c^{2} a + b^{2} c + c^{2} b}$

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 $\color{#3D99F6}{x+y =z}$ .

Since the inequality is symmetric, WLOG we can assume that $\color{#20A900}{z \ge x \ge y}$ .

Now substituting $x+y = z$ in the inequality , we get $\color{#D61F06}{ 2x^{2}(2y+x) + 2y^{2}(2x+y) + 2(x+y)^{3} \geq x^{3} + y^{3} + (x+y)^{3} + 9xyz}$

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

which is true by Rearrangement Inequality

Harsh Shrivastava - 5 years, 3 months ago

Can you write the equality cases for Case 2?

Sharky Kesa - 5 years, 3 months ago

I don't think equality exists . . .

Problem 8

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

Daniel Liu - 5 years, 3 months 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

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

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.

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

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

Sualeh Asif - 5 years, 3 months ago

Problem 6 :

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

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

Problem 13

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

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.

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

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

Problem 5

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

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

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.

Problem 14

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

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

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

Chris Galanis - 5 years, 3 months ago

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

A Former Brilliant Member - 5 years, 3 months ago

@A Former Brilliant Member – 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

Problem 25

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

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

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

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

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

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

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 - 5 years, 2 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 ;)

Daniel Liu - 5 years, 2 months ago

PROBLEM 2 :

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

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

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.

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.

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

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

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

Department 8 - 5 years, 3 months 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.

Problem 10

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

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

PROBLEM 12

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

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

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

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

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 $x, y$ .

@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.

Department 8 - 5 years, 1 month ago

Sure!

Harsh Shrivastava - 5 years, 1 month ago

More than welcome :)

A Former Brilliant Member - 5 years, 1 month ago

Waiting forit to begin...

Sualeh Asif - 5 years, 1 month ago

Problem 11 :

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

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

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$