<s>[INST] Prove that $\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{a+b+c}{2}$ If $a,b,c$ are positive real numbers, not all equal, prove that  $$\dfrac{bc}{b+c}+\dfrac{ca}{c+a}+\dfrac{ab}{a+b}<\dfrac{a+b+c}{2}$$
Progress:
$\frac{bc}{b+c}-\frac{b+c}{2}=$
 [/INST] Since $(b+c)^2 \geq 4bc$, we have
$$\dfrac{bc}{b+c} \leq \dfrac{b+c}4$$
Use the same inequality for the rest and obtain what you want.
</s>