Inequality for minimum value

let us take a+b+c+d = x

then the above will be ====>

= $\frac{x-(b+c+d)}{b+c+d}$ + $\frac{x-(a+c+d)}{a+c+d}$ + $\frac{x-(a+b+d)}{a+b+d}$ + $\frac{x-(a+b+c)}{a+b+c}$ + $\frac{x-a}{a}$ + $\frac{x-b}{b}$ + $\frac{x-c}{c}$ + $\frac{x-d}{d}$

On further simplification :-

= -4 + (x)( $\frac{1}{b+c+d}$ + $\frac{1}{b+c+d}$ + $\frac{1}{b+c+d}$ + $\frac{1}{b+c+d}$ ) -4 +(x)( $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ + $\frac{1}{d}$ ) ----- A

on applying AM-HM inequality on $\frac{1}{a}$ , $\frac{1}{b}$ , $\frac{1}{c}$ , $\frac{1}{d}$ and on $\frac{1}{b+c+d}$ , $\frac{1}{a+c+d}$ , $\frac{1}{a+b+d}$ , $\frac{1}{a+b+c}$ WE GET ====>

$\sum_{cyc}$ $\frac{1}{a}$ > $\frac{16}{x}$ ------- B

$\sum_{cyc}$ $\frac{1}{b+c+d}$ > $\frac{16}{3x}$ -------- C

on adding B and C and subtracting 8 we get A's MINIMUM :-

A_{_min_} = -8 + 16 + $\frac{16}{3}$ = $\frac{40}{3}$ = 13.33