A final homogeneous inequality

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0 replies

jlacosta

Mar 2, 2025

0 replies

About old Inequality

perfect_square 0

4 minutes ago

Source: Arqady

This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $\frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $uvw$ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$\frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}}$
Is there an another way?

0 replies

1 viewing

perfect_square

4 minutes ago

0 replies

inquality

karasuno 1

N 27 minutes ago by sqing

The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$ . Prove the inequality $(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$

1 reply

1 viewing

karasuno

an hour ago

sqing

27 minutes ago

Number Theory

karasuno 0

an hour ago

Solve the equation $n!+10^{2014}=m^{4}$ in natural numbers m and n.

0 replies

karasuno

an hour ago

0 replies

Minimum number of values in the union of sets

bnumbertheory 5

N an hour ago by Rohit-2006

Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3

For each positive integer $n$ , let $f(n)$ denote the smallest possible value of $|A_1 \cup A_2 \cup \dots \cup A_n|$ where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$ . Determine $f(n)$ for each positive integer $n$ .

5 replies

bnumbertheory

Oct 14, 2023

Rohit-2006

an hour ago

No more topics!

A final homogeneous inequality

v_Enhance 1

N Jul 26, 2013 by sirknightingfail

Source: ELMO Shortlist 2013: Problem A9, by David Stoner

Let $a, b, c$ be positive reals, and let $\sqrt[2013]{\frac{3}{a^{2013}+b^{2013}+c^{2013}}}=P$ . Prove that
$\prod_{\text{cyc}}\left(\frac{(2P+\frac{1}{2a+b})(2P+\frac{1}{a+2b})}{(2P+\frac{1}{a+b+c})^2}\right)\ge \prod_{\text{cyc}}\left(\frac{(P+\frac{1}{4a+b+c})(P+\frac{1}{3b+3c})}{(P+\frac{1}{3a+2b+c})(P+\frac{1}{3a+b+2c})}\right).$ Proposed by David Stoner

1 reply

v_Enhance

Jul 23, 2013

sirknightingfail

Jul 26, 2013

A final homogeneous inequality

G H J

Reveal topic tags

inequalities

function

logarithms

inequalities unsolved

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Source: ELMO Shortlist 2013: Problem A9, by David Stoner

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v_Enhance

6857 posts

v_Enhance

#1 h Jul 23, 2013, 1:24 AM • 4 Y

Y by megarnie, HamstPan38825, Rounak_iitr, Adventure10

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sirknightingfail

260 posts

sirknightingfail

#2 h Jul 26, 2013, 6:09 PM • 4 Y

Y by AFSA, megarnie, Adventure10, Mango247

solution

Let us consider the function $f(x)=\ln(P+\frac{1}{x})$
Then we have that $f'(x)=\frac{-1}{Px+x^2}$
If $i>j>0$ then $(P+i+j)(i-j)>0$ , so $Pi+i^2>Pj+j^2$ .
Taking the negative reciprocal gets that
$\frac{-1}{Pi+i^2}>\frac{-1}{Pj+j^2}$ for all $i>j$
Therefore $f$ is convex on $(0,\infty)$
Therefore, for all $x,y>0$ , the statement $\frac{1}{2}(f(x)+f(y))\ge f(\frac{x+y}{2})$ is true.
Plugging in $(a+2b+3c,3a+2b+c),(4a+2b,4b+2a),(4a+2b,4a+2c),$ and $(a+2b+3c,3a+2b+c)$ for $(x,y)$ then summing results in the following:

$f(a+2b+3c)+f(3a+2b+2c)+f(4a+2b)+\frac{1}{2}(f(4a+2c)+f(4b+2a))\ge 2f(2(a+b+c))+f(3a+3b)+f(4a+b+c)$

subtracting $f(3a+2b+2c)+f(a+2b+3c)+2f(2(a+b+c))$ from both sides results in the following:

$f(4a+2b)+\frac{1}{2}(f(4a+2c)+f(4b+2a))\ge f(3a+3b)+f(4a+b+c)-f(3a+2b+c)-f(a+2b+3c)$

taking the cyclic sum of this, we get the following:

$\sum_{\text{cyc}} (f(4a+2b)+f(2a+4b)-2f(2(a+b+c))) \ge \sum_{\text{cyc}} (f(3b+3c)+f(4a+b+c)-f(3a+2b+c)-f(3a+b+2c))$
Exponentiating and plugging in the definition of $f$ gets use the following:

$\prod_{\text{cyc}} (\frac{(P+\frac{1}{4a+2b})(P+\frac{1}{2a+4b})}{(P+\frac{1}{2(a+b+c)})^2}\ge \prod_{\text{cyc}}(\frac{(P+\frac{1}{4a+b+c})(P+\frac{1}{3b+3c})}{(P+\frac{1}{3a+2b+c})(P+\frac{1}{3a+b+2c})})$
Multiplying the top and bottom by 4 inside the product on the left hand side gets the following:

$\prod_{\text{cyc}}(\frac{(2P+\frac{1}{2a+b})(2P+\frac{1}{a+2b})}{(2P+\frac{1}{a+b+c})^2})\ge \prod_{\text{cyc}}(\frac{(P+\frac{1}{4a+b+c})(P+\frac{1}{3b+3c})}{(P+\frac{1}{3a+2b+c})(P+\frac{1}{3a+b+2c})})$

Q.E.D.

Note: P only needs to be nonnegative for this inequality to hold.

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