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Difference between revisions of "2004 USAMO Problems/Problem 5"

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(Alternate solution. Previous alternate solution is incorrect.)
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(''Titu Andreescu'')
 
(''Titu Andreescu'')
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Let <math> \displaystyle a </math>, <math> \displaystyle b </math>, and <math> \displaystyle c </math> be positive real numbers.  Prove that
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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>
  
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This follows from [[Hölder's Inequality]]
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We present two proofs of this inequality:
  
 +
* By [[Hölder's Inequality]],
 +
 +
<center>
 +
<math>
 +
\begin{matrix}(m_{1,1} + m_{1,2} + m_{1,3})(m_{2,1} + m_{2,2} + m_{2,3})(m_{3,1} + m_{3,2} + m_{3,3}) \ge \\ \qquad \qquad \qquad\left[ (m_{1,1}m_{2,1}m_{3,1})^{1/3} + (m_{2,1}m_{2,2}m_{2,3})^{1/3} + (m_{3,1}m_{3,2}m_{3,3})^{1/3} \right] ^3 \end{matrix}.
 +
</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>
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(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.
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</math>
 +
</center>
 +
Applying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get
 +
<center>
 +
<math>
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3(x^3 + y^3 + z^3) \ge (x^2 + y^2 + z^2)(x+y+z),
 +
</math>
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</center>
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and
 +
<center>
 +
<math>
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(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>
−
\begin{matrix}(m_{1,1} + m_{1,2} + m_{1,3})(m_{2,1} + m_{2,2} + m_{2,3})(m_{3,1} + m_{3,2} + m_{3,3}) \ge \\ \qquad \qquad \qquad\left[ (m_{1,1}m_{2,1}m_{3,1})^{1/3} + (m_{2,1}m_{2,2}m_{2,3})^{1/3} + (m_{3,1}m_{3,2}m_{3,3})^{1/3} \right] ^3 \end{matrix}
+
(x^6 + 2)(y^6 + 2)(z^6 + 2) \ge (x^2 + y^2 + z^2)^3,
 
</math>
 
</math>
 
</center>
 
</center>
−
when we take <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>.
+
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 $a$, $b$, and $c$ be positive real numbers. Prove that

$(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3$.

Solutions

We first note that for positive $x$, $x^5 + 1 \ge x^3 + x^2$. We may prove this in the following ways:

  • Since $x^2 - 1$ and $x^3 - 1$ have the same sign, $0 \le (x^2 - 1)(x^3 - 1) = x^5 - x^3 - x^2 + 1$, with equality when $x = 1$.
  • By weighted AM-GM, $\frac{2}{5}x^5 + \frac{3}{5} \ge x^2$ and $\frac{3}{5}x^5 + \frac{2}{5} \ge x^3$. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of Muirhead's Inequality.

It thus becomes sufficient to prove that

$(a^3 + 2)(b^3 + 2)(c^3 + 2) \ge (a+b+c)^3$.

We present two proofs of this inequality:

$\begin{matrix}(m_{1,1} + m_{1,2} + m_{1,3})(m_{2,1} + m_{2,2} + m_{2,3})(m_{3,1} + m_{3,2} + m_{3,3}) \ge \\ \qquad \qquad \qquad\left[ (m_{1,1}m_{2,1}m_{3,1})^{1/3} + (m_{2,1}m_{2,2}m_{2,3})^{1/3} + (m_{3,1}m_{3,2}m_{3,3})^{1/3} \right] ^3 \end{matrix}.$

We get the desired inequality by taking $m_{1,1} = a^3$, $m_{2,2} = b^3$, $m_{3,3} = c^3$, and $m_{x,y} = 1$ when $x \neq y$. We have equality if and only if $a = b = c = 1$.

  • Take $x = \sqrt{a}$, $y = \sqrt{b}$, and $z = \sqrt{c}$. Then some two of $x$, $y$, and $z$ are both at least $1$ or both at most $1$. Without loss of generality, say these are $x$ and $y$. Then the sequences $(x, 1, 1)$ and $(1, 1, y)$ are oppositely sorted, yielding

$(x^6 + 1 + 1)(1 + 1 + y^6) \ge 3(x^6 + 1 + y^6)$

by Chebyshev's Inequality. By the Cauchy-Schwarz Inequality we have

$(x^6 + 1 + y^6)(1 + z^6 + 1) \ge (x^3 + y^3 + z^3)^2.$

Applying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get

$3(x^3 + y^3 + z^3) \ge (x^2 + y^2 + z^2)(x+y+z),$

and

$(x^3 + y^3 + z^3)(x+y+z) \ge (x^2 + y^2 + z^2)^2.$

Multiplying the above four inequalities together yields

$(x^6 + 2)(y^6 + 2)(z^6 + 2) \ge (x^2 + y^2 + z^2)^3,$

as desired, with equality if and only if $x = y = z = 1$.

It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that $x^5 - x^2 + 3 \ge x^3 + 2$.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

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