Algebraic proof of a simple inequality

Question

Prove $\frac{a+b+c}{2} ≥ \frac{bc}{b+c} + \frac{ca}{c+a} + \frac{ab}{a+b}$

I feel like this problem would have a much simpler solution than just expaning out and simplifying. Any pointers are appreciated

Answer 1

Hint $$bc\leq {(b+c)^2\over 4}$$

Answer 2

By C-S $$\sum_{cyc}\frac{ab}{a+b}=\sum_{cyc}\frac{1}{\frac{1}{a}+\frac{1}{b}}\leq\sum_{cyc}\frac{1}{\frac{(1+1)^2}{a+b}}=\sum_{cyc}\frac{a+b}{4}=\frac{a+b+c}{2}.$$

Another way:

It's $$\sum_{cyc}\frac{ab}{a+b}\leq\frac{a+b+c}{2}$$ or $$\sum_{cyc}\left(\frac{a+b}{4}-\frac{ab}{a+b}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2}{a+b}\geq0.$$ Done!