Difference between revisions of "2004 USAMO Problems/Problem 5"
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== Solutions == | == Solutions == | ||
− | We first note that for positive <math> | + | We first note that for positive <math>x </math>, <math> x^5 + 1 \ge x^3 + x^2 </math>. We may prove this in the following ways: |
− | * Since <math> | + | * Since <math>x^2 </math> and <math>x^3 </math> must be both lesser than, both equal to, or both greater than 1, by the [[rearrangement inequality]], <math> x^2 \cdot x^3 + 1 \cdot 1 \ge x^2 \cdot 1 + 1 \cdot x^3 </math>. |
− | * Since <math> | + | * Since <math>x^2 - 1 </math> and <math>x^3 - 1 </math> have the same sign, <math> 0 \le (x^2 - 1)(x^3 - 1) = x^5 - x^3 - x^2 + 1 </math>, with equality when <math>x = 1 </math>. |
* By weighted [[AM-GM]], <math> \frac{2}{5}x^5 + \frac{3}{5} \ge x^2 </math> and <math> \frac{3}{5}x^5 + \frac{2}{5} \ge x^3 </math>. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of [[Muirhead's Inequality]]. | * By weighted [[AM-GM]], <math> \frac{2}{5}x^5 + \frac{3}{5} \ge x^2 </math> and <math> \frac{3}{5}x^5 + \frac{2}{5} \ge x^3 </math>. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of [[Muirhead's Inequality]]. | ||
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<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 | + | |
</math>. | </math>. | ||
</center> | </center> | ||
− | Setting <math> | + | Setting <math>a = m_{1,1} </math>, <math>b = m_{2,2} </math>, <math>c = m_{3,3} </math>, and <math>m_{x,y} = 1 </math> when <math> x \neq y </math> gives us the desired inequality, with equality when <math>x = y = z = 1 </math>. |
Second, we may apply the [[Cauchy-Schwarz Inequality]] twice to obtain | Second, we may apply the [[Cauchy-Schwarz Inequality]] twice to obtain | ||
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<math> | <math> | ||
\begin{matrix} | \begin{matrix} | ||
− | + | \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 \; \; | & \ge & (a + b + c)^4 \qquad \qquad \quad \; \; | ||
\end{matrix} | \end{matrix} | ||
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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> | + | ''Unfortunately, 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 23:02, 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.
First, Hölder's Inequality gives us
.
Setting , , , and when gives us the desired inequality, with equality when .
Second, we may apply the Cauchy-Schwarz Inequality twice to obtain
,
as desired.
Unfortunately, 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.