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Difference between revisions of "2004 USAMO Problems/Problem 5"
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== Solutions ==
 
== Solutions ==
  
−
We first note that for positive <math> \displaystyle x </math>, <math> x^5 + 1 \ge x^3 + x^2 </math>.  We may prove this in the following ways:
+
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> \displaystyle x^2 </math> and <math> \displaystyle 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>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> \displaystyle x^2 - 1 </math> and <math> \displaystyle 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> \displaystyle x = 2 </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 \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}
+
\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>.
 
</math>.
 
</center>
 
</center>
−
Setting <math> \displaystyle a = m_{1,1} </math>, <math> \displaystyle b = m_{2,2} </math>, <math> \displaystyle c = m_{3,3} </math>, and <math> \displaystyle m_{x,y} = 1 </math> when <math> x \neq y </math> gives us the desired inequality, with equality when <math> \displaystyle x = y = z = 1 </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}
−
\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 \\
+
\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> \displaystyle x^5 - x^2 + 3 \ge x^3 + 2 </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 $\displaystyle a$, $\displaystyle b$, 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 $x$, $x^5 + 1 \ge x^3 + x^2$. 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.

First, Hölder's Inequality gives us

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

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

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

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

as desired.


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


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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