During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

2014 USAJMO Problems/Problem 1

Problem

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

Solution

Since $(a-1)^5\ge 0$, $$a^5-5a^4+10a^3-10a^2+5a-1\ge 0$$ or $$10a^2-5a+1\le a^3(a^2-5a+10)$$ Since $a^2-5a+10=\left( a-\dfrac{5}{2}\right)^2 +\dfrac{15}{4}>0$, $$\frac{10a^2-5a+1}{a^2-5a+10}\le a^3$$ Also note that $10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}\ge 0$, We conclude $$0\le \frac{10a^2-5a+1}{a^2-5a+10}\le a^3$$ Similarly, $$0\le \frac{10b^2-5b+1}{b^2-5b+10}\le b^3$$ $$0\le \frac{10c^2-5c+1}{c^2-5c+10}\le c^3$$ So $$\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\le a^3b^3c^3$$ or $$\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \le(abc)^3$$ Therefore, $$\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\le abc.$$