Question 96
Prove that
$$
\frac{3}{2} \lt \frac{4a+b}{a+4b} + \frac{4b+c}{b+4c} + \frac{4c+a}{c+4a} \lt 9
$$
for positive real numbers $a, b, c$.
$$
\frac{3}{2} \lt \frac{4a+b}{a+4b} + \frac{4b+c}{b+4c} + \frac{4c+a}{c+4a} \lt 9
$$
for positive real numbers $a, b, c$.