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

 
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== Problem 5 ==
 
== Problem 5 ==
  
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(''Titu Andreescu'') Let <math> \displaystyle a </math>, <math> \displaystyle b </math>, and <math> \displaystyle c </math> be positive real numbers.  Prove that
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(''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
 
<center>
 
<center>
 
<math>
 
<math>
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as desired.
 
as desired.
  
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Unfortunately, it is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that <math> \displaystyle x^5 - x^2 + 3 \ge x^3 + 2 </math>.
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''Unfortunately, it is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that <math> \displaystyle x^5 - x^2 + 3 \ge x^3 + 2 </math>.''
  
  

Revision as of 22:10, 24 March 2007

Problem 5

(Titu Andreescu) Let $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$ 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 $\displaystyle x$, $x^5 + 1 \ge x^3 + x^2$. We may prove this in the following ways:

  • Since $\displaystyle x^2 - 1$ and $\displaystyle x^3 - 1$ have the same sign, $0 \le (x^2 - 1)(x^3 - 1) = x^5 - x^3 - x^2 + 1$, with equality when $\displaystyle x = 2$.
  • 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.

First, Hölder's Inequality gives us

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

Setting $\displaystyle a = m_{1,1}$, $\displaystyle b = m_{2,2}$, $\displaystyle c = m_{3,3}$, and $\displaystyle m_{x,y} = 1$ when $x \neq y$ gives us the desired inequality, with equality when $\displaystyle x = y = z = 1$.

Second, we may apply the Cauchy-Schwarz Inequality twice to obtain

$\begin{matrix} \displaystyle \left[(a^2 + 1 + 1)(1 + b^2 + 1)\right] \left[ (1 + 1 + c^2)(a + b + c) \right] & \ge & (a + b + 1)^2( a + b + c^2)^2 \\ & \ge & (a + b + c)^4 \qquad \qquad \quad \; \; \end{matrix}$,

as desired.


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


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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