Let the roots in question be $x=2+p, 3-q$ (for $p,q \ge 0$) such that $2+p \le 3-q \Rightarrow p+q \le 1$. This gives us the feasible region $\Omega =\{(p, q): p+q \le 1; p, q \ge 0\}$ with extreme points $(0,0); (1,0); (0,1)$. For our quadratic equation we have: $x^2 - \frac{4b}{a}x + \frac{4c}{a} = [x - (2+p)][x-(3-q)] = 0 \Rightarrow \frac{4b}{a} = 5 + p -q; \frac{4c}{a} = 6+3p-2q-pq$; or $b = \frac{5+p-q}{4}a; c = \frac{6+3p-2q-pq}{4}a$ (i). Part (a): Let us verify the inequality $a \le b \le c < a+b$ holds via the extreme points $(p,q) \in \Omega$: (1) $(p,q) = (0,0): (a,b,c) = \left(a, \frac{5}{4}a, \frac{3}{2}a \right)$ and $a + \frac{5}{4}a > \frac{3}{2}a \Rightarrow \frac{9}{4} > \frac{3}{2}$; (2) $(p,q) = (1,0): (a,b,c) = \left(a, \frac{3}{2}a, \frac{9}{4}a \right)$ and $a + \frac{3}{2}a > \frac{9}{4}a \Rightarrow \frac{5}{2} > \frac{9}{4}$; (3) $(p,q) = (0,1): (a,b,c) = \left(a, a, a \right)$ and $a + a > a \Rightarrow 2 > 1$; which all 3 cases are true for $a > 0$. Part (b): Let us verify the inequality $\frac{a}{a+c}+\frac{b}{b+a} > \frac{c}{b+c}$ holds via the extreme ordered triples computed in Part (a): (1) $(a,b,c) = \left(a, \frac{5}{4}a, \frac{3}{2}a \right) \Rightarrow \frac{a}{a+3a/2} + \frac{5a/4}{5a/4+a} > \frac{3a/2}{3a/2+5a/4} \Rightarrow \frac{43}{45}>\frac{6}{11}$; (2) $(a,b,c) = \left(a, \frac{3}{2}a, \frac{9}{4}a \right) \Rightarrow \frac{a}{a+9a/4}+\frac{3a/2}{3a/2+a}>\frac{9a/4}{9a/4+3a/2} \Rightarrow \frac{59}{65} > \frac{3}{5}$; (3) $(a,b,c) = \left(a, a, a \right) \Rightarrow \frac{a}{a+a}+\frac{a}{a+a}>\frac{a}{a+a} \Rightarrow 2 > 1$; which all 3 cases are true for $a > 0$. QED