<s>[INST] 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.
 [/INST] You can indeed use AM-GM, as the book says, to derive the result. Admittedly, it would have been more transparent had the book added one line of derivation as shown below.
For positive $x$ and $y$,
$$\frac{x+y}2\,\frac12\Big(\frac1x+\frac1y\Big)\ge\sqrt{xy}\sqrt{\frac1{xy}}=1,$$
thus
$$\frac1x+\frac1y\ge \frac4{x+y}.$$
Substituing $x=b+c$ and $y=a+d$, you get the desired inequality.
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