Difference between revisions of "2004 USAMO Problems/Problem 5"
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− | Let <math> | + | Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers. Prove that |
<center> | <center> | ||
<math> | <math> | ||
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</center> | </center> | ||
− | + | We present two proofs of this inequality: | |
+ | * By [[Hölder's Inequality]], | ||
+ | |||
+ | <center> | ||
+ | <math> | ||
+ | |||
+ | </math> | ||
+ | </center> | ||
+ | We get the desired inequality by taking <math>m_{1,1} = a^3</math>, <math>m_{2,2} = b^3</math>, <math>m_{3,3} = c^3</math>, and <math>m_{x,y} = 1 </math> when <math> x \neq y </math>. We have equality if and only if <math>a = b = c = 1 </math>. | ||
+ | |||
+ | * Take <math>x = \sqrt{a}</math>, <math>y = \sqrt{b}</math>, and <math>z = \sqrt{c}</math>. Then some two of <math>x</math>, <math>y</math>, and <math>z</math> are both at least <math>1</math> or both at most <math>1</math>. Without loss of generality, say these are <math>x</math> and <math>y</math>. Then the sequences <math>(x, 1, 1)</math> and <math>(1, 1, y)</math> are oppositely sorted, yielding | ||
+ | <center> | ||
+ | <math> | ||
+ | (x^6 + 1 + 1)(1 + 1 + y^6) \ge 3(x^6 + 1 + y^6) | ||
+ | </math> | ||
+ | </center> | ||
+ | by [[Chebyshev's Inequality]]. By the [[Cauchy-Schwarz Inequality]] we have | ||
+ | <center> | ||
+ | <math> | ||
+ | (x^6 + 1 + y^6)(1 + z^6 + 1) \ge (x^3 + y^3 + z^3)^2. | ||
+ | </math> | ||
+ | </center> | ||
+ | Applying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get | ||
+ | <center> | ||
+ | <math> | ||
+ | 3(x^3 + y^3 + z^3) \ge (x^2 + y^2 + z^2)(x+y+z), | ||
+ | </math> | ||
+ | </center> | ||
+ | and | ||
+ | <center> | ||
+ | <math> | ||
+ | (x^3 + y^3 + z^3)(x+y+z) \ge (x^2 + y^2 + z^2)^2. | ||
+ | </math> | ||
+ | </center> | ||
+ | Multiplying the above four inequalities together yields | ||
<center> | <center> | ||
<math> | <math> | ||
− | + | (x^6 + 2)(y^6 + 2)(z^6 + 2) \ge (x^2 + y^2 + z^2)^3, | |
</math> | </math> | ||
</center> | </center> | ||
− | + | as desired, with equality if and only if <math>x = y = z = 1</math>. | |
''It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that <math>x^5 - x^2 + 3 \ge x^3 + 2 </math>.'' | ''It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that <math>x^5 - x^2 + 3 \ge x^3 + 2 </math>.'' |
Revision as of 23:53, 1 July 2008
Problem 5
(Titu Andreescu)
Let ,
, and
be positive real numbers. Prove that
.
Solutions
We first note that for positive ,
. We may prove this in the following ways:
- Since
and
must be both lesser than, both equal to, or both greater than 1, by the rearrangement inequality,
.
- Since
and
have the same sign,
, with equality when
.
- By weighted AM-GM,
and
. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of Muirhead's Inequality.
It thus becomes sufficient to prove that
.
We present two proofs of this inequality:
We get the desired inequality by taking ,
,
, and
when
. We have equality if and only if
.
- Take
,
, and
. Then some two of
,
, and
are both at least
or both at most
. Without loss of generality, say these are
and
. Then the sequences
and
are oppositely sorted, yielding
by Chebyshev's Inequality. By the Cauchy-Schwarz Inequality we have
Applying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get
and
Multiplying the above four inequalities together yields
as desired, with equality if and only if .
It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.