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

Revision as of 12:40, 24 July 2009

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.

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

@@ Line 4: / Line 4: @@
 ==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>
 Apply [[AM-GM]] to get
@@ Line 12: / Line 10: @@
 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>
-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}}