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Difference between revisions of "2014 USAJMO Problems/Problem 1"

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==Problem==
 
==Problem==
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Let <math>a</math>, <math>b</math>, <math>c</math> be real numbers greater than or equal to <math>1</math>. Prove that <cmath>\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc </cmath>
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Let <math>a</math>, <math>b</math>, <math>c</math> be real numbers greater than or equal to <math>1</math>. Prove that <cmath>\min{\left (\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc </cmath>
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==Solution==
 
==Solution==
 
Since <math>(a-1)^5\geqslant 0</math>,
 
Since <math>(a-1)^5\geqslant 0</math>,

Revision as of 15:17, 23 July 2014

Problem

Let $a$, $b$, $c$ be real numbers greater than or equal to $1$. Prove that \[\min{\left (\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc\]

Solution

Since $(a-1)^5\geqslant 0$, \[a^5-5a^4+10a^3-10a^2+5a-1\geqslant 0\] or \[10a^2-5a+1\leqslant a^3(a^2-5a+10)\] Since $a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}\geqslant 0$, \[\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3\] Also note that $10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}\geqslant 0$, We conclude \[0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3\] Similarly, \[0\leqslant\frac{10b^2-5b+1}{b^2-5b+10}\leqslant b^3\] \[0\leqslant\frac{10c^2-5c+1}{c^2-5c+10}\leqslant c^3\] So \[\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\leqslant a^3b^3c^3\] or \[\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \leqslant(abc)^3\] Therefore, \[\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc.\]

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