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?
 A: HINT: set 
$$b+c=x$$
$$c+a=y$$
$$a+b=z$$
adding we get
$$a+b+c=\frac{x+y+z}{2}$$ and we can compute $$a+x=\frac{x+y+z}{2}$$ thus $$a=\frac{-x+y+z}{2}$$ etc
A: The first step of the solution says that $X+\frac1X+Y+\frac1Y+Z+\frac1Z\geq2+2+2$.
where $X=\frac{a+b}{b+c}$ and so on.
Do you know that $X+\frac1X$ is always $2$ or more whenever $X>0$?
A: 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
A: More easier and obvious answer:
$$\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\geq\frac{3}{2}$$
$$\frac{2a}{b+c}+\frac{2c}{a+b}+\frac{2b}{a+c}\geq3$$
$$\frac{2a+b+c}{b+c}+\frac{b+2c+a}{a+b}+\frac{c+2b+a}{a+c}\geq6$$
$$\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq6$$
