Difference between revisions of "1974 USAMO Problems/Problem 2"

(Solution)
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==Solution==
 
==Solution==
Taking the natural log of both sides, we obtain
+
Consider the function <math>f(x)=x\ln{x}</math>. <math>f''(x)=\frac{1}{x}>0</math> for <math>x>0</math>; therefore, it is a convex function and we can apply [[Jensen's Inequality]]:
<center><math>a\ln{a}+b\ln{b}+c\ln{c}\ge \left(\frac{a+b+c}{3}\right)\ln(abc)</math></center>
 
It is sufficient to prove the above inequality. Consider the function <math>f(x)=x\ln{x}</math>. <math>f''(x)=\frac{1}{x}>0</math> for <math>x>0</math>; therefore, it is a convex function and we can apply [[Jensen's Inequality]]:
 
 
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)</math></center>
 
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)</math></center>
 
Apply [[AM-GM]] to get
 
Apply [[AM-GM]] to get
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which implies
 
which implies
 
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt[3]{abc}\right)</math></center>
 
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt[3]{abc}\right)</math></center>
which is equivalent to the desired inequality.
+
Rearranging,
 +
<center><math>a\ln{a}+b\ln{b}+c\ln{c}\ge\left(\frac{a+b+c}{3}\right)\ln\left(abc\right)</math></center>
 +
Because <math>f(x) = e^x</math> is an increasing function, we can conclude that:
 +
<center><math>e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}</math></center>
 +
which simplifies to the desired inequality.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 12:40, 24 July 2009

Problem

Prove that if $a$, $b$, and $c$ are positive real numbers, then

$a^ab^bc^c\ge (abc)^{(a+b+c)/3}$

Solution

Consider the function $f(x)=x\ln{x}$. $f''(x)=\frac{1}{x}>0$ for $x>0$; therefore, it is a convex function and we can apply Jensen's Inequality:

$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)$

Apply AM-GM to get

$\frac{a+b+c}{3}\ge \sqrt[3]{abc}$

which implies

$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt[3]{abc}\right)$

Rearranging,

$a\ln{a}+b\ln{b}+c\ln{c}\ge\left(\frac{a+b+c}{3}\right)\ln\left(abc\right)$

Because $f(x) = e^x$ is an increasing function, we can conclude that:

$e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}$

which simplifies to the desired inequality.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1974 USAMO (ProblemsResources)
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