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

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 $10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}> 0$, 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, \[\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\le abc.\]

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