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 $10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)^2+\dfrac{3}{8}> 0$ , 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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2014 USAJMO Problems/Problem 1

Problem

Solution