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 . 

By C-S we obtain:
$$\frac{ab}{a+b}+\frac{cd}{c+d}=a+c+\left(\frac{ab}{a+b}-a\right)+\left(\frac{cd}{c+d}-c\right)=$$
$$=a+c-\left(\frac{a^2}{a+b}+\frac{c^2}{c+d}\right)\leq a+c-\frac{(a+c)^2}{a+b+c+d}=\frac{(a+c)(b+d)}{a+b+c+d}.$$
Done!
