<s>[INST] Prove that $\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$ $$\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$$
I tried applying a.m. g.m inequality to l.h.s and tried to find upper bound for l.h.s and lower bound for r.h.s but i am not getting answer . 
 [/INST] I suppose that $a,b,c,d$ are $>0$. An idea is to put
$$F(x)=\frac{(x+c)(b+d)}{x+b+c+d}-\frac{xb}{x+b}-\frac{cd}{c+d}$$
and to compute the derivative:
$$F^{\prime}(x)=\frac{(b+d)^2}{(x+b+c+d)^2}-\frac{b^2}{(x+b)^2}$$
This show that the minimum of $F$ on $]0,+\infty[$ is obtained for $x=bc/d$, and to finish you have to compute this minimum. 
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