Quadratic equation with tricky conditions. Need to prove resulting inequalities. 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

As shown in other answers, let $\alpha,\beta$ be the two real roots of $ax^2-bx+c=0$, i.e. $$ax^2-bx+c=a(x-\alpha)(x-\beta)\iff \alpha+\beta=\frac{b}{a},\ \alpha\beta=\frac{c}{a}.$$
Since $\alpha,\beta\in[2,2.4]$ and $a>0$,
$$\frac{b}{a}=\alpha+\beta\ge 4\Rightarrow a\le \frac{b}{4}<b;$$ 
$$\frac{b}{c}=\frac{\alpha+\beta}{\alpha\beta}=\frac{1}{\alpha}+\frac{1}{\beta}\le 1\Rightarrow b\le c;$$
and
$$(\alpha-1)(\beta-1)<2\Rightarrow 1+\frac{b}{a}=\alpha+\beta+1>\alpha\beta=\frac{c}{a}\Rightarrow c< a+b.$$
Therefore,
$$\frac{a}{a+c}+\frac{b}{a+b}\ge \frac{a}{a+c}+\frac{b}{a+c}=\frac{a+b}{a+c}>\frac{c}{a+c},$$
where the first inequality is due to $a,b>0$ and $0\le b\le c$ and the second one is due to $a+b>c$ and $a+c>0$.
