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

(New page: ==Problem== Prove that if <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers, then <center><math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math></center> ==Solution== {{so...)
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
−
{{solution}}
+
Taking the natural log of both sides, we obtain
 +
<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
 +
<center><math>\frac{a+b+c}{3}\ge \sqrt[3]{abc}</math></center>
 +
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.
  
−
==See also==
+
{{alternate solutions}}
 +
 
 +
== Mathlinks Discussions ==
 +
*[http://www.mathlinks.ro/viewtopic.php?t=102633 Simple Olympiad Inequality]
 +
*[http://www.mathlinks.ro/viewtopic.php?t=98846 Hard inequality]
 +
*[http://www.mathlinks.ro/viewtopic.php?t=85663 Inequality]
 +
*[http://www.mathlinks.ro/Forum/viewtopic.php?t=82706 Some q's on usamo write ups]
 +
*[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}}
 
{{USAMO box|year=1974|num-b=1|num-a=3}}
  
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]

Revision as of 00:47, 3 January 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

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
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