<s>[INST] Prove that $\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b} \leq \frac{a+b+c}{2}$ If $a,b,c \in R$, then prove that:
$$\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b} \leq \frac{a+b+c}{2}$$
I can't see any known inequality working here like $H.M.-A.M.$. Could this be solved using basic inequalities?
 [/INST] For $a,b,c>0$ we get
$(a-b)^2=(a+b)^2-4ab\geq 0\implies \frac{ab}{a+b}\leq \frac{a+b}{4}$, the equality occurs only when $a=b$.  
In similar way we have
$\frac{bc}{b+c}\leq \frac{b+c}{4}$ and $\frac{ca}{c+a}\leq \frac{c+a}{4}$
Adding these we get 
$$\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b} \leq \frac{2(a+b+c)}{4}=\frac{a+b+c}{2}$$
The equality occurs only when $a=b=c$ 
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