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}\geqslant 0</math>, | Since <math>a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}\geqslant 0</math>, | ||
<cmath> \frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3 </cmath> | <cmath> \frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3 </cmath> | ||
− | Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)+\dfrac{3}{8}\geqslant 0</math>, | + | Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}\geqslant 0</math>, |
We conclude | We conclude | ||
<cmath>0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3</cmath> | <cmath>0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3</cmath> |
Revision as of 22:52, 29 April 2014
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,