If `V` is the volume of a cuboid of dimensions `a , b , ca n dS` is its surface area, then prove that `S/V = 2(1/a+1/b+1/c)`

Explanation

To prove
`1/v=2/s[1/a+1/b+1/c]`
Consider LHS
`1/v=1/(abc)............(i)`
Consider RHS
`2/s[1/a+1/b+1/c]`
`2/(2(ab+bc+ac))[1/a+1/b+1/c]`
`1/(ab+bc+ac)[1/a+1/b+1/c]`
`1/(ab+bc+ac)[(ab+bc+ac)/(abc)]`
`1/(abc)`
`2/s[1/a+1/b+1/c]=1/(abc)................(ii)`
Hence from (i) and (ii)
we get `1/v=2/s[1/a+1/b+1/c]`