<s>[INST] prove that $\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$ Let a,b,c,d be positive real numbers such that$\ a+b+c+d=1$, than prove that $$\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$$
This question is from the practice material of Indian Math Olympiad. I think that this will incorporate the AM-GM-HM inequality.I am not able to learn the perfect heuristic to solve this inequality. 
 [/INST] By AM-GM:
$ab\le \left(\frac{a+b}{2}\right)^2=\frac{(a+b)^2}{4}$
Thus
$$\sum_{\text{sym}}\frac{ab}{a+b}\le \sum_{\text{sym}}\frac{a+b}{4}=\frac{3}{4}$$
Equality is achieved iff $a=b=c=d$.
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