<s>[INST] Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of solution given is $\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq2+2+2=6$
I don't know how to proceed from the question to the first step of solution. Can anyone explain?
 [/INST] We have three expressions:
$$S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$
$$M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}$$
$$N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$$
Obviously, we have $M+N=3$ According to AM-GM,we have:
$$M+S=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}\geqslant 3$$
$$N+S=\frac{a+c}{b+c}+\frac{a+b}{c+a}+\frac{b+c}{a+b}\geqslant 3$$
Thus, $M+N+2S\geqslant 6$ so $2S\geqslant 3$ so $S\geqslant \dfrac{3}{2}$ q.e.d
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