# 2008 iTest Problems/Problem 90

## Problem

For $a,b,c$ positive reals, let $N=\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{c^2+a^2}{b^2+ca}$. Find the minimum value of $\lfloor 2008N\rfloor$.

## Solution 1

By the Trivial Inequality (with equality happening if $a = b$), \begin{align*} a^2 - 2ab + b^2 &\ge 0 \\ a^2 + b^2 &\ge 2ab. \end{align*} Add $2c^2$ to both sides and use the reciprocal property to get \begin{align*} a^2 + b^2 + 2c^2 &\ge 2ab + 2c^2 \\ \frac{1}{2ab+2c^2} &\ge \frac{1}{a^2 + b^2 + 2c^2}. \end{align*} Since $2a^2 + 2b^2 \ge 0$, multiplying both sides by this value would not change the inequality sign, and doing so results in \begin{align*} \frac{a^2+b^2}{ab+c^2} &\ge \frac{2a^2 + 2b^2}{a^2+b^2+2c^2} \\ &\ge 2 \left( \frac{a^2+b^2}{a^2+c^2+b^2+c^2} \right). \end{align*} By using similar steps, we find that $N \ge 2 \left( \frac{a^2+b^2}{a^2+c^2+b^2+c^2} + \frac{b^2+c^2}{b^2+a^2+c^2+a^2} + \frac{a^2+c^2}{a^2+b^2+c^2+b^2}\right).$

Let $x = a^2+b^2$, $y = b^2+c^2$, and $z = a^2+c^2$, making $N \ge 2\left(\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y}\right)$. Note that $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} = (x+y+z)(\frac{1}{y+z} + \frac{1}{x+z} + \frac{1}{x+y}) - 3$. By the Cauchy-Schwarz Inequality, $(y+z+x+z+x+y)(\frac{1}{y+z} + \frac{1}{x+z} + \frac{1}{x+y}) \ge 9$, so $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} \ge \frac32$. Equality happens if $(y+z)^2 = (x+z)^2 = (x+y)^2$, which is possible if $x = y = z$. If $x = y = z$, then $a = b = c$.

Therefore, the minimum value of $N$ (which happens if $a = b = c$) is $2 \cdot \frac32 = 3$, so the minimum value of $\lfloor 2008N\rfloor$ is $\boxed{6024}$.