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

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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> | 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> | ||

==Solution== | ==Solution== | ||

+ | Notice <math>\dfrac{10a^2 - 5a + 1}{a^2 - 5a + 10} \le a^3</math> rearranges to <math>(a-1)^5 \ge 0</math>, obvious. Therefore <cmath> \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 ) \le (abc)^3 </cmath> so <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> | ||

+ | == Headline text == |

## Revision as of 17:33, 29 April 2014

## Problem

Let , , be real numbers greater than or equal to . Prove that

## Solution

Notice rearranges to , obvious. Therefore so