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!
 A: 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.
A: Hints: $\times (a+b)(c+d)(a+b+c+d)$ on the left side and right side.
A: The solution of @Jack D'Aurizio  is a bit brutal. I would like to show a more tricky solution.
You can prove the next inequality Cauchy-Schwarz inequality:
$$   \sum\limits_{i=1}^n\frac{x_i^2}{y_i}\geq\frac{(\sum\limits_{i=1}^nx_i)^2}{\sum\limits_{i=1}^ny_i},             $$
which is simple for people who has experience in Math Olympics.
So, by simple transformation, we can show that the inequality in the title is equivalent to:
\begin{align}
&\frac{2ab}{a+b}+\frac{2cd}{c+d}≤\frac{2(a+c)(b+d)}{a+b+c+d} \\
\iff &\frac{(a+b)^2-(a^2+b^2)}{a+b}+\frac{(c+d)^2-(c^2+d^2)}{c+d}≤\frac{(a+b+c+d)^2-[(a+c)^2+(b+d)^2]}{a+b+c+d} \\
\iff &\frac{(a+b)^2}{a+b}+\frac{(c+d)^2}{c+d}+\frac{(a+c)^2+(b+d)^2}{a+b+c+d} ≤\frac{a^2+b^2}{a+b}+\frac{ c^2+d^2 }{c+d}+\frac{(a+b+c+d)^2}{a+b+c+d} \\
\iff & \frac{(a+c)^2+(b+d)^2}{a+b+c+d} ≤\frac{a^2+b^2}{a+b}+\frac{ c^2+d^2 }{c+d}
\end{align}
Notice that:
$$ \frac{(a+c)^2}{a+b+c+d}\leq  \frac{a^2}{a+b}+\frac{ c^2}{c+d}, \frac{(b+d)^2}{a+b+c+d}\leq  \frac{b^2}{a+b}+\frac{ d^2}{c+d},$$
which proves this inequality.
