2014 USAJMO Problems/Problem 1

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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