Difference between revisions of "1974 USAMO Problems/Problem 2"
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Note that <math>a^a \ge a</math> even if <math>a \le 1</math>. Therefore <math>a^a \ge a</math>, <math>b^b \ge b</math>, and <math>c^c \ge c</math>. Multiplying these together we get: | Note that <math>a^a \ge a</math> even if <math>a \le 1</math>. Therefore <math>a^a \ge a</math>, <math>b^b \ge b</math>, and <math>c^c \ge c</math>. Multiplying these together we get: | ||
<math>a^ab^bc^c \ge abc = 1</math>. This proves the desired result. Equality holds when <math>a = b = c = 1</math>. | <math>a^ab^bc^c \ge abc = 1</math>. This proves the desired result. Equality holds when <math>a = b = c = 1</math>. | ||
+ | |||
+ | ==Solution 8 (Rearrangement)== | ||
+ | Let <math>\log_2a=x</math>, <math>\log_2b=y</math> and <math>\log_2c=z</math>, and WLOG <math>a\ge b\ge c >0</math>. | ||
+ | Then we have both <math>a\ge b\ge c</math> and <math>x\ge y \ge z</math>. | ||
+ | By the rearrangement inequality, | ||
+ | <math>ax+by+cz\ge ay + bz +cx</math> and | ||
+ | <math>ax+by +cz\ge az+bx+cy</math>. | ||
+ | Summing, | ||
+ | <math>2(ax+by+cz)\ge a(y+z)+b(x+z)+c(x+y)</math> | ||
+ | Adding <math>ax+by+cz</math>,we get | ||
+ | <math>ax+by+cz\ge \frac{(a+b+c)(x+y+z)}{3}</math>. | ||
+ | Now we substitute back for <math>x,y,z</math> to get: | ||
+ | <math>\log_2a^ab^bc^c\ge \log_2(abc)^{\frac{a+b+c}{3}}</math>. | ||
+ | Raising <math>2</math> to the power of each side, we get | ||
+ | <math>a^ab^bc^c\ge (abc)^{\frac{a+b+c}{3}}</math> | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 07:21, 30 May 2020
Contents
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
.
Solution 5
Let
and
Then
and a straightforward calculation reduces the problem to
WLOG, assume
Then
and
Therefore,
J.Z.
Solution 6
Cubing both sides of the given inequality gives
If we take
as the product of
's,
's, and
, we get that
by GM-HM, as desired.
Solution 7
Replacing with
,
with
, and
with
, for some positive real
we get:
This means that this inequality is homogeneous since both sides have the same power of
as a factor. Since the inequality is homogeneous, we can scale
so that their product is
, i.e.
. This makes the inequality turn into something much more nicer to deal with. Now we have to prove:
given that
.
Note that
even if
. Therefore
,
, and
. Multiplying these together we get:
. This proves the desired result. Equality holds when
.
Solution 8 (Rearrangement)
Let ,
and
, and WLOG
.
Then we have both
and
.
By the rearrangement inequality,
and
.
Summing,
Adding
,we get
.
Now we substitute back for
to get:
.
Raising
to the power of each side, we get
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.