Difference between revisions of "1974 USAMO Problems/Problem 2"
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This is clearly true for <math>k \ge 1</math>. | This is clearly true for <math>k \ge 1</math>. | ||
+ | ==Solution 4== | ||
+ | WLOG let <math>a\ge b\ge c</math>. Then sequence <math>(a,b,c)</math> majorizes <math>(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})</math>. Thus by Muirhead's Inequality, we have <math>\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}</math>, so <math>a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}</math>. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 23:41, 11 May 2018
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 1
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)
Rearranging,
![$a\ln{a}+b\ln{b}+c\ln{c}\ge\left(\frac{a+b+c}{3}\right)\ln\left(abc\right)$](http://latex.artofproblemsolving.com/3/f/5/3f59079c18ed4396607cf01df15d21c4dcb3d272.png)
Because is an increasing function, we can conclude that:
![$e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}$](http://latex.artofproblemsolving.com/6/3/5/635872e58e16d76df675269c7674dff286ad4dd5.png)
which simplifies to the desired inequality.
Solution 2
Note that .
So if we can prove that and
, then we are done.
WLOG let .
Note that . Since
,
,
, and
, it follows that
.
Note that . Since
,
,
, and
, it follows that
.
Thus, , and cube-rooting both sides gives
as desired.
Solution 3
WLOG let . Let
and
, where
and
.
We want to prove that .
Simplifying and combining terms on each side, we get .
Since , we can divide out
to get
.
Take the th root of each side and then cube both sides to get
.
This simplifies to .
Since and
, we only need to prove
for our given
.
WLOG, let and
for
. Then our expression becomes
This is clearly true for .
Solution 4
WLOG let . Then sequence
majorizes
. Thus by Muirhead's Inequality, we have
, so
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1974 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
- Simple Olympiad Inequality
- Hard inequality
- Inequality
- Some q's on usamo write ups
- ineq
- exponents (generalization)
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.