Nesbitt's Inequality for 4 Variables

I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.

Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negative real number, the following inequality holds:
$$\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}\ge 2$$

I have no idea how the author proves this. It comes under the very first section, AM-GM. I get the original Nesbitt's inequality in 3 variables that the author proves (which is also cryptic, but I was able to decipher it).

My effort: I understood how the author defines the variables $M, N$ and $S$.
$$S = \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}$$
$$M = \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b}$$

$$N = \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b}$$

$M + N = 4$, pretty straightforward. The numerators and denominators cross out to give four 1s.

Then the author, without any expansion/explanation, says

$$M + S = \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b}\ge 4$$

Which is also true, since the AM-GM inequality says

$$\frac{M+S}{4}\ge \left(\frac{a+b}{b+c}\cdot\frac{b+c}{c+d}\cdot\frac{c+d}{d+a}\cdot\frac{d+a}{a+b}\right)^{1/4}$$

The RHS above evaluates to $1^{1/4}$ since all the numerators and denominators cancel out.

The next part is the crux of my question.

The author claims,

$$N + S =\frac{a+c}{b+c}+\frac{a+c}{a+d}+\frac{b+d}{c+d} + \frac{b+d}{a+b}\ge\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}$$

This is completely bizarre for me! Where did the author manage to get a sum of $(a+b+c+d)$??

As a side note, I'd definitely not recommend this book for any beginner in basic algebraic inequalities (even though the title of the book promotes that it's a treatment of basic inequalities). The author takes certain 'leaps of faith', just assuming that the student reading the book would be able to follow.

Answer

Since we have $(x-y)^2\ge 0$, we have, for $x\gt 0,y\gt 0$,
$$\begin{align}(x-y)^2\ge 0&\Rightarrow x^2+y^2+2xy\ge 4xy\\&\Rightarrow y(x+y)+x(x+y)\ge 4xy\\&\Rightarrow \frac{1}{x}+\frac 1y\ge\frac{4}{x+y}\end{align}$$
Now set $x=b+c,y=a+d$ and $x=c+d,y=a+b$ to get
$$\frac{1}{b+c}+\frac{1}{a+d}\ge\frac{4}{b+c+a+d}$$and$$\frac{1}{c+d}+\frac{1}{a+b}\ge\frac{4}{c+d+a+b}.$$