Prove Inequality: $\sqrt{ab}+\sqrt{cd}\le \sqrt{(a+d)(b+c)}$

[anemone]

Prove that for positive reals $a,\,b,\,c,\,d$, $\sqrt{ab}+\sqrt{cd}\le \sqrt{(a+d)(b+c)}$.

 

[lfdahl]

Spoiler

We have:

$$\sqrt{ab} + \sqrt{cd} \leq \sqrt{(a+d)(b+c)} \\\\
\left ( \sqrt{ab} + \sqrt{cd}\right )^2 \leq \left ( \sqrt{ab + ac + bd + cd} \right )^2
\\\\ab + cd + 2\sqrt{ab}\sqrt{cd}\leq ab + ac + bd + cd
\\\\ac + bd - 2\sqrt{ab}\sqrt{cd}\geq 0
\\\\\left ( \sqrt{ac} \right )^2 + \left ( \sqrt{bd} \right )^2-2\sqrt{ac}\sqrt{bd}\geq 0
\\\\\left ( \sqrt{ac}-\sqrt{bd} \right )^2 \geq 0. $$

Thus the inequality holds.