The roots of the quadratic equation $ax^ 2-bx+c=0,$ $a>0$, both lie within the interval $[2,\frac{12}{5}]$. Prove that: (a) $a \leq b \leq c <a+b$. (b) $\frac{a}{a+c}+\frac{b}{b+a}>\frac{c}{c+b}$
So we can use the quadractic formula and obtain $2 \leq \frac{b \pm \sqrt{b^2-4ac}}{2a} \leq \frac{12}{5}$
and as $a>0$, this implies that
$4a \leq b \pm \sqrt{b^2-4ac} \leq \frac{24a}{5}$.
Then we can divide this into two cases (plus or minus) --- But how can we manipulate it to yield the required result?
Another idea is to make use of the fact that $b^2-4ac$ is non-negative