TITLE: Given $a > b > c > d > 0,$ and U $= \sqrt{ab} + \sqrt{cd}$ , V $= \sqrt{ac} + \sqrt{bd}$ , W $= \sqrt{ad} + \sqrt{bc}$.
QUESTION [5 upvotes]: Given $a > b > c > d > 0,$ and U $= \sqrt{ab} + \sqrt{cd}$, $V = \sqrt{ac} + \sqrt{bd}$, $ W = \sqrt{ad} + \sqrt{bc}$, arrange $U$,$V$,$W$ in ascending order.

What I Tried: I squared $U$,$V$,$W$ to get :-
$\rightarrow U^2 = ab + cd + 2\sqrt{abcd}$. 
$\rightarrow V^2 = ac + bd + 2\sqrt{abcd}$. 
$\rightarrow  W^2 = ad + bc + 2\sqrt{abcd}$.
We can cancel out the $2\sqrt{abcd}$ from each and we are only left to compare $(ab + cd) , (ac + bd) , (ad + bc)$ in order to compare $U$$,V$,$W$.
This is where I get stuck. I could say that $ab > ac$ , but I couldn't necessarily show that $cd > bd$ , as $b > c$. That way does not work for me, and I am not finding any other way to compare these.
Can anyone help?

REPLY [4 votes]: Hint: With your second & third expressions you're comparing, we get
$$\begin{equation}\begin{aligned}
(ac + bd) - (ad + bc) & = ac - ad + bd - bc \\
& = a(c - d) + b(d - c) \\
& = a(c - d) - b(c - d) \\
& = (a - b)(c - d) \\
& \gt 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$