2014 USAJMO Problems/Problem 1: Difference between revisions
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==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> | 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> | ||
Revision as of 17:35, 29 April 2014
Problem
Let 
, 
, 
be real numbers greater than or equal to 
. Prove that ![\[\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc\]](http://latex.artofproblemsolving.com/3/a/d/3ad6f9ab816469889e5b8452f203c0e9b01647f4.png)
Solution
Notice 
rearranges to 
, obvious. Therefore ![\[\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right ) \le (abc)^3\]](http://latex.artofproblemsolving.com/c/5/6/c56e6c136df2f0c6e6e5c3175121d5385a66bfeb.png)
so ![\[\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc.\]](http://latex.artofproblemsolving.com/5/8/6/586aa2ebfc9ff2507752ab79760d05d894abe799.png)
