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?

More easier and obvious answer:
$$\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\geq\frac{3}{2}$$
$$\frac{2a}{b+c}+\frac{2c}{a+b}+\frac{2b}{a+c}\geq3$$
$$\frac{2a+b+c}{b+c}+\frac{b+2c+a}{a+b}+\frac{c+2b+a}{a+c}\geq6$$
$$\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq6$$
