<s>[INST] Algebraic Inequality involving AM-GM-HM If $$a,b,c \;\epsilon \;R^+$$ 
Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$\frac{bc}{x-a} + \frac{ab}{x-c} + \frac{ac}{x-b} \leq \frac{1}{2} \Bigl(x\Bigl)$$ 
This didn't help so instead I wrote it as $$2\Biggl(\frac {1}{a} + \frac{1}{b} + \frac{1}{c}\Biggl ) \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
Not sure what to to do next.
Any help would be appreciated.
 [/INST] Multiply both sides by $4$ and rearrange:
$$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl) \Rightarrow \\
\left(a+b-\frac{4ab}{a+b}\right)+\left(b+c-\frac{4bc}{b+c}\right)+\left(c+a-\frac{4ca}{c+a}\right)\ge 0 \Rightarrow \\
\frac{(a-b)^2}{a+b}+\frac{(b-c)^2}{b+c}+\frac{(c-a)^2}{c+a}\ge 0.$$
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