2014 USAJMO Problems/Problem 1

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 $\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc.$

Page

Toolbox

Search

2014 USAJMO Problems/Problem 1

Problem

Solution