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- Feb 14, 2012

- 3,802

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

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,802

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

- Nov 26, 2013

- 732

$$\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.