Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$

Question

Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$ My steps

A= $\frac{a}{b+c}$,B= $\frac{b}{c+a}$ &C=$\frac{c}{a+b}$

A.M$\ge$H.M

$\frac{A+B+C}{3} \ge \frac{3ABC}{AB+BC+AC}$

I am struck after this step

${A+B+C} \ge \frac{9}{\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}}$

Answer 1

This is Nesbitt's inequality. You find many proofs of that classic one. See here.

Answer 2

Hint: \begin{eqnarray*} \sum_{cyc} (a+b)(a-b)^2 \geq 0 \end{eqnarray*} This can be rearranged to \begin{eqnarray*} 2(a(a+b)(a+c)+b(b+c)(b+a)+c(c+a)(c+b))-3(a+b)(a+c)(b+c) \geq 0 \end{eqnarray*}

Answer 3

set $$x=b+c,y=c+a$$ and $$z=a+b$$ then we get $$\frac{-x+y+z}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$ and this is $$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-3\geq 3 $$ and then use $$\frac{A}{B}+\frac{B}{A}\geq 2$$ for $$A,B>0$$

Answer 4

$$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{2a-b-c}{(b+c)}=$$ $$= \frac{1}{2}\sum_{cyc}\frac{a-b-(c-a)}{(b+c)}=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{1}{b+c}-\frac{1}{a+c}\right)=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{(a+c)(b+c)}\geq0.$$ Done!

Another way.

By C-S $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2}.$$