Nesbitt? No!

Using re-arrangement inequality,

$\frac{a}{S-a} + \frac{b}{S-b} + \frac{c}{S-c} + \frac{d}{S-d} \geq \frac{a}{S-b} + \frac{b}{S-c} + \frac{c}{S-d} + \frac{d}{S-a}$ where S represents a + b + c + d

Similarly, keeping it in all the permutations and then adding,

$n(\frac{a}{S-a} + \frac{b}{S-b} + \frac{c}{S-c} + \frac{d}{S-d}) \geq (a + b + c + d) (\frac{1}{S-b} + \frac{1}{S-c} + \frac{1}{S-d} + \frac{1}{S-a})$

Now using Cauchy Schwarz in Engel form,

$\frac{1}{S-b} + \frac{1}{S-c} + \frac{1}{S-d} + \frac{1}{S-a} \geq \frac{16}{4S-S}$

Substituting,

$\frac{a}{S-a} + \frac{b}{S-b} + \frac{c}{S-c} + \frac{d}{S-d} \geq \frac{16S}{12S}$

$\geq \frac{4}{3}$

Well, here we draw a conclusion that $\frac{{a}_{1}}{S-{a}_{1}} + \frac{{a}_{2}}{S-{a}_{2}} + \frac{{a}_{3}}{S-{a}_{3}} +...... + \frac{{a}_{n}}{S-{a}_{n}} \geq \frac{n+1}{n}$ for any $n \in {Z}^{+}$

We can use Jensen's inequality. Since the equation is homogeneous, we can assume that a+b+c+d=1. So the expression becomes a/(1-a)+b/(1-b)+c/(1-c)+d/(1-d). Now consider f(x)=x/(1-x). The second derivative of this is 2/(1-x)^3 which is positive as 0<x<1 This means that the function is convex.

So we know that 1/4((a/(1-a)+b/(1-b)+c/(1-c)+d/(1-d))>= f(1/4)=1/3.

Multiply by 4 to get the required answer.

Aside: We can generalise this argument where the denominator has n terms to get that the sum is >= n+1/n .

The problem is the $n = 4$ case of the generalised nestbitt's inequality .

Call the expression P. We have $S=a(b+c+d)+b(c+d+a)+c(d+a+b)+d(a+b+c)=2[(a+d)(b+c)+ad+bc]$ Using Cauchy-Schwarz Inequality we get $P.S\geq (a+b+c+d)^2$ By AM-GM, we get $\frac{S}{2}\leq \frac{(a+d)^2}{4}+\frac{(b+c)^2}{4}+\frac{(a+d)(b+c)}{2}+\frac{(a+d)(b+c)}{2}\leq \frac{(a+b+c+d)^2}{4}+\frac{(a+b+c+d)^2}{8}$ $\Rightarrow S\leq\frac{3}{4}(a+b+c+d)^2$ $\therefore P\geq\frac{4}{3}$ The equality holds when $a=b=c=d$

Since interchanging a, b, c, d does not change the expression, if all are equal, we get extreme value. Since the expression has > in it, the expression with a=b=c=d could be the maximum. Thus we get 1.333. I am not sure my logic is correct.