If V is the volume of a cuboid of dimensions a, b, c and S is its surface area,

It is given that $V$ is the volume of a cuboid of length $=a$, breadth $=b$ and height $=c$. Also, $S$ is surface area of cuboid.

Then, $V=a \times b \times c$

Surface area of the cuboid $=2 \times($ length $\times$ breadth $+$ breadth $\times$ height $+$ length $\times$ height $)$

$\Rightarrow S=2 \times(a \times b+b \times c+a \times c)$

Let us take the right - hand side of the equation to be proven.

$\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{2}{2 \times(a \times b+b \times c+a \times c)} \times\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

$=\frac{1}{(a \times b+b \times c+a \times c)} \times\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

Now, multiplying the numerator and the denominator with $a \times b \times c$, we get:

$\frac{1}{(a \times b+b \times c+a \times c)} \times\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \times \frac{a \times b \times c}{a \times b \times c}$

$=\frac{1}{(a \times b+b \times c+a \times c)} \times\left(\frac{a \times b \times c}{a}+\frac{a \times b \times c}{b}+\frac{a \times b \times c}{c}\right) \times \frac{1}{a \times b \times c}$

$=\frac{1}{(a \times b+b \times c+a \times c)} \times(b \times c+a \times c+a \times b) \times \frac{1}{a \times b \times c}$

$=\frac{1}{(a \times b+b \times c+a \times c)} \times(a \times b+b \times c+a \times c) \times \frac{1}{a \times b \times c}$

$=\frac{1}{a \times b \times c}$

$=\frac{1}{V}$

$\therefore \frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{V}$