Difference between revisions of "1997 USAMO Problems/Problem 5"
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Prove that, for all positive real numbers <math>a, b, c,</math> | Prove that, for all positive real numbers <math>a, b, c,</math> | ||
− | <math> | + | <math>\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}</math>. |
== Solution 1 == | == Solution 1 == | ||
Line 18: | Line 18: | ||
== Solution 2 == | == Solution 2 == | ||
Rearranging the AM-HM inequality, we get <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}</math>. Letting <math>x=a^{3}+b^{3}+abc</math>, <math>y=b^{3}+c^{3}+abc</math>, and <math>z=c^{3}+a^{3}+abc</math>, we get <cmath>\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc}.</cmath> By AM-GM on <math>a^{3}</math>, <math>b^{3}</math>, and <math>c^{3}</math>, we have <cmath>a^{3}+b^{3}+c^{3} \ge 3abc \Rightarrow 2a^{3}+2b^{3}+2c^{3}+3abc \ge 9abc \Rightarrow \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc} \le \frac{1}{abc}.</cmath> So, <math>\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{1}{abc}</math>. | Rearranging the AM-HM inequality, we get <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}</math>. Letting <math>x=a^{3}+b^{3}+abc</math>, <math>y=b^{3}+c^{3}+abc</math>, and <math>z=c^{3}+a^{3}+abc</math>, we get <cmath>\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc}.</cmath> By AM-GM on <math>a^{3}</math>, <math>b^{3}</math>, and <math>c^{3}</math>, we have <cmath>a^{3}+b^{3}+c^{3} \ge 3abc \Rightarrow 2a^{3}+2b^{3}+2c^{3}+3abc \ge 9abc \Rightarrow \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc} \le \frac{1}{abc}.</cmath> So, <math>\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{1}{abc}</math>. | ||
+ | -Tigerzhang | ||
− | - | + | <math>\textbf{WARNING:}</math> |
+ | |||
+ | This solution doesn’t work because <math>(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 9</math>, so <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | If we multiply each side by <math>abc</math>, we get that we must just prove that <cmath> \frac{abc}{a^3+b^3+abc} + \frac{abc}{b^3+c^3+abc} + \frac{abc}{c^3+a^3+abc} \leq 1</cmath> If we divide our LHS equation, we get that <cmath>\frac{1}{\frac{a^3+b^3+abc}{abc}} +\frac{1}{\frac{b^3+c^3+abc}{abc}} + \frac{1}{\frac{c^3+a^3+abc}{abc}}</cmath> <cmath> = \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} +\frac{1}{\frac{b^2}{ac} + \frac{c^2}{ab} + 1} + \frac{1}{\frac{c^2}{ab} + \frac{a^2}{bc} + 1} </cmath> Make the astute observation that by Titu's Lemma, <cmath> \frac{a^2}{bc} + \frac{b^2}{ac} \geq \frac{\left(a+b\right)^2}{bc+ ac}</cmath> <cmath> \sum_{cyc}\frac{\left(a+b\right)^2}{bc+ ac} \leq \sum_{cyc}\frac{a^2}{bc} + \frac{b^2}{ac} </cmath> Therefore: <cmath> \sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} \geq \sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1}</cmath> If we expand it out, we get that <cmath> \sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} = \frac{a+b+c}{a+b+c} = 1</cmath> Since our original equation is less than this, we get that <cmath> \sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} \leq 1</cmath> <cmath> \sum_{cyc} \frac{abc}{a^3+b^3+abc}\leq 1</cmath> <cmath> \sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \frac{1}{abc}</cmath> | ||
+ | |||
+ | -KEVIN_LIU | ||
==See Also == | ==See Also == |
Latest revision as of 13:55, 25 March 2024
Problem
Prove that, for all positive real numbers
.
Solution 1
Because the inequality is homogenous (i.e. can be replaced with
without changing the inequality other than by a factor of
for some
), without loss of generality, let
.
Lemma:
Proof: Rearranging gives
, which is a simple consequence of
and
Thus, by :
Solution 2
Rearranging the AM-HM inequality, we get . Letting
,
, and
, we get
By AM-GM on
,
, and
, we have
So,
.
-Tigerzhang
This solution doesn’t work because , so
Solution 3
If we multiply each side by , we get that we must just prove that
If we divide our LHS equation, we get that
Make the astute observation that by Titu's Lemma,
Therefore:
If we expand it out, we get that
Since our original equation is less than this, we get that
-KEVIN_LIU
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.