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...) |
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==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. | ||
− | == | + | {{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 01:47, 3 January 2009
Problem
Prove that if ,
, and
are positive real numbers, then
![$a^ab^bc^c\ge (abc)^{(a+b+c)/3}$](http://latex.artofproblemsolving.com/c/d/b/cdb94a39e3190f57fcbb1f2fbe849a6c038b66e7.png)
Solution
Taking the natural log of both sides, we obtain
![$a\ln{a}+b\ln{b}+c\ln{c}\ge \left(\frac{a+b+c}{3}\right)\ln(abc)$](http://latex.artofproblemsolving.com/b/c/7/bc763f15d16af4b6324b045d8338d22a0fc06b54.png)
It is sufficient to prove the above inequality. Consider the function .
for
; therefore, it is a convex function and we can apply Jensen's Inequality:
![$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)$](http://latex.artofproblemsolving.com/d/b/0/db0d87edcf1329bb08f89821fd547f050f23f363.png)
Apply AM-GM to get
![$\frac{a+b+c}{3}\ge \sqrt[3]{abc}$](http://latex.artofproblemsolving.com/8/7/2/872ec383ca01e8997916c1dbaee116fee66f8130.png)
which implies
![$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt[3]{abc}\right)$](http://latex.artofproblemsolving.com/d/a/3/da34b15dd7c8d75509c763b00ecc8f379c8cbd22.png)
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
- Simple Olympiad Inequality
- Hard inequality
- Inequality
- Some q's on usamo write ups
- ineq
- exponents (generalization)
1974 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |