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2004 USAMO Problems/Problem 5: Difference between revisions

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(''Titu Andreescu'')
(''Titu Andreescu'')
Let <math> \displaystyle a </math>, <math> \displaystyle b </math>, and <math> \displaystyle c </math> be positive real numbers.  Prove that
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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== Solutions ==
== Solutions ==
https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat]


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>


We present two proofs of this inequality.
We present two proofs of this inequality:
 
* By [[Hölder's Inequality]],


First, [[Hölder's Inequality]] gives us
<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>.
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>.


Second, we may apply the [[Cauchy-Schwarz Inequality]] twice to obtain
* 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>
\begin{matrix}
(x^6 + 2)(y^6 + 2)(z^6 + 2) \ge (x^2 + y^2 + z^2)^3,
\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 \\
</math>
& \ge & (a + b + c)^4 \qquad \qquad \quad \; \;
</center>
\end{matrix}
as desired, with equality if and only if <math>x = y = z = 1</math>.
</math>,
* First, expand the left side of the inequality to get
<center>
<math>a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8.</math>
</center>
By the AM-GM inequality, it is true that <math>a^3 + a^3b^3 + 1 \ge 3a^2b</math>, and so it is clear that
<center>
<math>a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8 \ge a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2.</math>
</center>
Additionally, again by AM-GM, it is true that <math>a^3b^3c^3 + a^3 + b^3 + c^3 + 1 + 1 \ge 6abc</math>, and so
<center>
<math>a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2</math>
</center>
</center>
<center>
<math>\ge a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2 + 6abc = {(a + b + c)}^3,</math>
</center>
as desired.
as desired.




''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>.''
 
''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>.''




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== Resources ==
== Resources ==


* [[2004 USAMO Problems]]
{{USAMO newbox|year=2004|num-b=4|num-a=6}}
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=17397#p17397 Discussion on AoPS/MathLinks]
 




[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Inequality Problems]]
{{MAA Notice}}

Latest revision as of 10:19, 26 September 2024

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

https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat]

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 $x = y = z = 1$.

  • First, expand the left side of the inequality to get

$a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8.$

By the AM-GM inequality, it is true that $a^3 + a^3b^3 + 1 \ge 3a^2b$, and so it is clear that

$a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8 \ge a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2.$

Additionally, again by AM-GM, it is true that $a^3b^3c^3 + a^3 + b^3 + c^3 + 1 + 1 \ge 6abc$, and so

$a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2$

$\ge a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2 + 6abc = {(a + b + c)}^3,$

as desired.


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

2004 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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