The sides of a triangle are a = 4, b = 6 and c = 8.

Given: $a=4, b=6$ and $c=8$.

Then,

$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{16+64-36}{2 \times 4 \times 8}=\frac{11}{16}$

$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{36+64-16}{2 \times 6 \times 8}=\frac{7}{8}$

$\cos C=\frac{b^{2}+a^{2}-c^{2}}{2 a b}=\frac{16+36-64}{2 \times 4 \times 6}=\frac{-1}{4}$

Now,

$8 \cos A+16 \cos B+4 \cos C=8 \times \frac{7}{8}+16 \times \frac{11}{16}-4 \times \frac{1}{4}$

$\Rightarrow 8 \cos A+16 \cos B+4 \cos C=7+11-1=17$

Hence proved.