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

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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>
== Headline text ==
 

Revision as of 17:35, 29 April 2014

Problem

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

Solution

Notice $\dfrac{10a^2 - 5a + 1}{a^2 - 5a + 10} \le a^3$ rearranges to $(a-1)^5 \ge 0$, obvious. Therefore \[\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right ) \le (abc)^3\] so \[\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc.\]