1974 USAMO Problems/Problem 2

Problem

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

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

Solution

Taking the natural log of both sides, we obtain

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

It is sufficient to prove the above inequality. 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

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

which is equivalent 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.

Mathlinks Discussions

1974 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Page

Toolbox

Search

1974 USAMO Problems/Problem 2

Problem

Solution

Mathlinks Discussions