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

Revision as of 13:56, 17 September 2012

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

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

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.

@@ Line 18: / Line 18: @@
 {{alternate solutions}}
-== Mathlinks Discussions ==
+== See Also ==
+{{USAMO box|year=1974|num-b=1|num-a=3}}
 *[http://www.mathlinks.ro/viewtopic.php?t=102633 Simple Olympiad Inequality]
 *[http://www.mathlinks.ro/viewtopic.php?t=98846 Hard inequality]
@@ Line 25: / Line 26: @@
 *[http://www.mathlinks.ro/viewtopic.php?t=213258 ineq]
 *[http://www.mathlinks.ro/Forum/viewtopic.php?t=46247 exponents (generalization)]
-{{USAMO box|year=1974|num-b=1|num-a=3}}
 [[Category:Olympiad Algebra Problems]]
 [[Category:Olympiad Inequality Problems]]

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