ACCORDING TO QUES,:

Sides of the triangle:

\[a\text{ }=\text{ }4,\text{ }b\text{ }=\text{ }6\text{ }and\text{ }c\text{ }=\text{ }8\]

By using the formulas,

\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]

\[Cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2ac\]

Or,

\[Cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2ab\]

Hence, let’s substitute the values of \[a,\text{ }b\text{ }and\text{ }c\]in the equation,

\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]

\[=\text{ }({{6}^{2}}~+\text{ }{{8}^{2}}~\text{ }{{4}^{2}})/2\times 6\times 8\]

Or,

\[=\text{ }\left( 36\text{ }+\text{ }64\text{ }\text{ }16 \right)/96\]

\[=\text{ }84/96\]

So,

\[=\text{ }7/8\]

Now,

\[Cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2ac\]

\[=\text{ }({{4}^{2}}~+\text{ }{{8}^{2}}~\text{ }{{6}^{2}})/2\times 4\times 8\]

Or,

\[=\text{ }\left( 16\text{ }+\text{ }64\text{ }\text{ }36 \right)/64\]

\[=\text{ }44/64\]

Now,

\[Cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2ab\]

\[=\text{ }({{4}^{2}}~+\text{ }{{6}^{2}}~\text{ }{{8}^{2}})/2\times 4\times 6\]

\[=\text{ }\left( 16\text{ }+\text{ }36\text{ }\text{ }64 \right)/48\]

Or,

\[=\text{ }-12/48\]

\[=\text{ }-1/4\]

So, considering LHS:

\[8\text{ }cos\text{ }A\text{ }+\text{ }16\text{ }cos\text{ }B\text{ }+\text{ }4\text{ }cos\text{ }C\]

\[=\text{ }8\text{ }\times \text{ }7/8\text{ }+\text{ }16\text{ }\times \text{ }44/64\text{ }+\text{ }4\text{ }\times \text{ }\left( -1/4 \right)\]

Or,

\[=\text{ }7\text{ }+\text{ }11\text{ }\text{ }1\]

\[=\text{ }17\]

Hence proved.