Show that if a,b,c,d are positive then $\frac{ab}{a+b}+\frac{cd}{c+d}\le \frac{(a+c)(b+d)}{a+b+c+d}$ Show that if a,b,c,d are positive then
$\frac{ab}{a+b}+\frac{cd}{c+d}\le \frac{(a+c)(b+d)}{a+b+c+d}$
I am stuck with this.
Thanks in advance!

The inequality is equivalent to:
$$(abc+abd+acd+bcd)(a+b+c+d)\leq(a+c)(b+d)(a+b)(c+d)\tag{1}$$
or to:
$$ b^2 c^2 + a^2 d^2\geq 2abcd \tag{2}$$
that is a consequence of the AM-GM inequality.
