Difference between revisions of "1974 USAMO Problems/Problem 2"
m (→Solution) |
Menuthreeone (talk | contribs) m (→Solution 9) |
||
(12 intermediate revisions by 10 users not shown) | |||
Line 3: | Line 3: | ||
<center><math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math></center> | <center><math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math></center> | ||
− | ==Solution== | + | ==Solution 1== |
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]]: | 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> | <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> | ||
Line 16: | Line 16: | ||
which simplifies to the desired inequality. | which simplifies to the desired inequality. | ||
+ | ==Solution 2== | ||
+ | Note that <math>(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>. | ||
+ | So if we can prove that <math>a^ab^bc^c\ge a^bb^cc^a</math> and <math>a^ab^bc^c\ge a^cb^ac^b</math>, then we are done. | ||
+ | WLOG let <math>a\ge b\ge c</math>. | ||
+ | |||
+ | Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a</math>. Since <math>\dfrac{c}{a} \le 1</math>, <math>\dfrac{c}{b} \le 1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^bb^cc^a</math>. | ||
+ | |||
+ | Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b</math>. Since <math>\dfrac{b}{a} \le 1</math>, <math>\dfrac{c}{a} \le 1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^cb^ac^b</math>. | ||
+ | |||
+ | Thus, <math>(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>, and cube-rooting both sides gives <math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math> as desired. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | WLOG let <math>a\ge b\ge c</math>. Let <math>b = ax</math> and <math>c = ay</math>, where <math>x \ge 1</math> and <math>y \ge 1</math>. | ||
+ | |||
+ | We want to prove that <math>(a)^{a}(ax)^{ax}(ay)^{ay} \ge (a \cdot ax \cdot ay)^{\frac{a + ax + ay}{3}}</math>. | ||
+ | |||
+ | Simplifying and combining terms on each side, we get <math>a^{a + ax + ay}x^{ax}y^{ay} \ge a^{a + ax + ay}(xy)^{\frac{a + ax + ay}{3}}</math>. | ||
+ | |||
+ | Since <math>a > 0</math>, we can divide out <math>a^{a + ax + ay}</math> to get <math>x^{ax}y^{ay} \ge (xy)^{\frac{a + ax + ay}{3}}</math>. | ||
+ | |||
+ | Take the <math>a</math>th root of each side and then cube both sides to get <math>x^{3x}y^{3y} \ge (xy)^{1 + x + y}</math>. | ||
+ | |||
+ | This simplifies to <math>x^{2x-1}y^{2y-1} \ge x^{y}y^{x}</math>. | ||
+ | |||
+ | Since <math>2x - 1 \ge x</math> and <math>2y - 1 \ge y</math>, we only need to prove <math>x^{x}y^{y} \ge x^{y}y^{x}</math> for our given <math>x, y</math>. | ||
+ | |||
+ | WLOG, let <math>y \ge x</math> and <math> y =kx</math> for <math>k \ge 1</math>. Then our expression becomes | ||
+ | |||
+ | <math>x^{x}(xk)^{xk} \ge x^{xk}(xk)^{x}</math> | ||
+ | |||
+ | <math>x^{x+xk}k^{xk} \ge x^{x+xk}k^{x}</math> | ||
+ | |||
+ | <math>k^{xk} \ge k^{x}</math> | ||
+ | |||
+ | <math>k^k \ge k</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>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>x=\frac{a}{\sqrt[3]{abc}},</math> <math>y=\frac{b}{\sqrt[3]{abc}}</math> and <math>z=\frac{c}{\sqrt[3]{abc}}.</math> Then <math>xyz=1</math> and a straightforward calculation reduces the problem to | ||
+ | <cmath>x^xy^yz^z \ge 1.</cmath> | ||
+ | WLOG, assume <math>x\ge y\ge z.</math> Then <math>x\ge 1,</math> <math>z\le 1</math> and <math>xy=\frac{1}{z} \ge 1.</math> Therefore, | ||
+ | <cmath> x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.</cmath> | ||
+ | |||
+ | J.Z. | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ==Solution 9== | ||
+ | Assume without loss of generality that <math>a\le b\le c</math>. Then, we have | ||
+ | <cmath>b/a\ge 1, c/a\ge 1, c/b\ge 1, b-a\ge 0, c-a\ge 0, c-b\ge 0.</cmath> | ||
+ | It follows that | ||
+ | <cmath>(b/a)^{(b-a)/3}\cdot (c/a)^{(c-a)/3}\cdot (c/b)^{(c-b)/3}\ge 1.</cmath> | ||
+ | We can "distribute" the exponents, which gives | ||
+ | <cmath>\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\cdot \dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\cdot \dfrac{c^{(c-b)/3}}{b^{(c-b)/3}}\ge 1.</cmath> | ||
+ | Notice that in the <math>b^{(c-b)/3}</math> in the denominator of the third fraction can be brought out of the denominator by negating its exponent. This gives | ||
+ | <cmath>\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\cdot \dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\cdot c^{(c-b)/3} \cdot b^{(b-c)/3}\ge 1.</cmath> | ||
+ | We can move around terms on the left to get | ||
+ | <cmath>\left(\dfrac{b^{(b-a)/3}\cdot b^{(b-c)/3}}{a^{(b-a)/3}}\right)\left(\dfrac{c^{(c-a)/3}\cdot c^{(c-b)/3}}{a^{(c-a)/3}}\right)\ge 1.</cmath> | ||
+ | Now we combine the exponents in the numerators. This gives | ||
+ | <cmath>\left(\dfrac{b^{(2b-c-a)/3}}{a^{(b-a)/3}}\right)\left(\dfrac{c^{(2c-b-a)/3}}{a^{(c-a)/3}}\right)\ge 1.</cmath> | ||
+ | We multiply both sides by <math>\left(a^{(b-a)/3}\right)\left(a^{(c-a)/3}\right),</math> which gives | ||
+ | <cmath>\left(b^{(2b-c-a)/3}\right)\left(c^{(2c-b-a)/3}\right)\ge \left(a^{(b-a)/3}\right)\left(a^{(c-a)/3}\right).</cmath> | ||
+ | We then multiply both sides by <math>\left(b^{(c-a)/3}\right)\left(c^{(b-a)/3}\right):</math> | ||
+ | <cmath>\left(b^{(2b-2a)/3}\right)\left(c^{(2c-2a)/3}\right)\ge \left(a^{(b-a)/3}\right)\left(c^{(b-a)/3}\right)\left(a^{(c-a)/3}\right)\left(b^{(c-a)/3}\right).</cmath> | ||
+ | Multiplying both sides by <math>\left(b^{(b-a)/3}\right)\left(c^{(c-a)/3}\right),</math> then simplifying both sides with exponent laws, gives | ||
+ | <cmath>b^{b-a}\cdot c^{c-a}\ge (abc)^{(b+c-2a)/3}.</cmath> | ||
+ | Finally, we multiply both sides by <math>(abc)^a.</math> Combining exponents one last time gives the desired | ||
+ | <cmath>a^a b^b c^c\ge (abc)^{(a+b+c)/3}.</cmath> | ||
+ | --MenuThreeOne | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Line 29: | Line 136: | ||
*[http://www.mathlinks.ro/viewtopic.php?t=213258 ineq] | *[http://www.mathlinks.ro/viewtopic.php?t=213258 ineq] | ||
*[http://www.mathlinks.ro/Forum/viewtopic.php?t=46247 exponents (generalization)] | *[http://www.mathlinks.ro/Forum/viewtopic.php?t=46247 exponents (generalization)] | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] |
Latest revision as of 14:01, 1 February 2024
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
Solution 9
Assume without loss of generality that . Then, we have
It follows that
We can "distribute" the exponents, which gives
Notice that in the
in the denominator of the third fraction can be brought out of the denominator by negating its exponent. This gives
We can move around terms on the left to get
Now we combine the exponents in the numerators. This gives
We multiply both sides by
which gives
We then multiply both sides by
Multiplying both sides by
then simplifying both sides with exponent laws, gives
Finally, we multiply both sides by
Combining exponents one last time gives the desired
--MenuThreeOne
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.