Difference between revisions of "1974 USAMO Problems/Problem 2"
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+ | ==Solution 6== | ||
+ | |||
+ | Cubing both sides of the given inequality gives <cmath>a^{3a}b^{3b}c^{3c}\ge(abc)^{a+b+c}</cmath> | ||
+ | If we take <math>a^{3a}b^{3b}c^{3c}</math> as the product of <math>3a</math> <math>a</math>'s, <math>3b</math> <math>b</math>'s, and <math>3c</math> <math>c's</math>, we get that | ||
+ | |||
+ | <center> | ||
+ | <math>\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\ge\frac{3(a+b+c)}{\frac{3a}{a}+\frac{3b}{b}+\frac{3c}{c}}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}</math> | ||
+ | </center> | ||
+ | |||
+ | by GM-HM, as desired. | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | Replacing <math>a</math> with <math>ak</math>, <math>b</math> with <math>bk</math>, and <math>c</math> with <math>ck</math>, for some positive real <math>k</math> we get: | ||
+ | <math>a^{ak}k^{ak}b^{bk}k^{bk}c^{ck}k^{ck} = a^{ak}b^{bk}c^{ck}k^{ak+bk+ck} \ge {k^3abc}^{{ak+bk+ck}/3} = k^{ak+bk+ck}{abc}^{{ak+bk+ck}/3}</math> | ||
+ | This means that this inequality is homogeneous since both sides have the same power of <math>k</math> as a factor. Since the inequality is homogeneous, we can scale <math>abc</math> so that their product is <math>1</math>, i.e. <math>abc = 1</math>. This makes the inequality turn into something much more nicer to deal with. Now we have to prove: | ||
+ | <math>a^ab^bc^c \ge 1</math> given that <math>abc = 1</math>. | ||
+ | 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>. | ||
+ | |||
+ | ==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}} |
Latest revision as of 06:21, 30 May 2020
Contents
Problem
Prove that if , , and are positive real numbers, then
Solution 1
Consider the function . for ; therefore, it is a convex function and we can apply Jensen's Inequality:
Apply AM-GM to get
which implies
Rearranging,
Because is an increasing function, we can conclude that:
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.