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

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Since <math>a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}>0</math>, | Since <math>a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}>0</math>, | ||

<cmath> \frac{10a^2-5a+1}{a^2-5a+10}\le a^3 </cmath> | <cmath> \frac{10a^2-5a+1}{a^2-5a+10}\le a^3 </cmath> | ||

− | Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8} | + | Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}> 0</math>, |

We conclude | We conclude | ||

<cmath>0\le \frac{10a^2-5a+1}{a^2-5a+10}\le a^3</cmath> | <cmath>0\le \frac{10a^2-5a+1}{a^2-5a+10}\le a^3</cmath> |

## Latest revision as of 20:23, 15 April 2018

## Problem

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

## Solution

Since , or Since , Also note that , We conclude Similarly, So or Therefore,