# Prove that $\sqrt{A_1A_2} + \sqrt{B_1B_2} \leq \sqrt{A_1 + B_1}\sqrt{A_2 + B_2}$

I am trying to prove the below inequality using the AM-GM inequality, but I can't see how to get it to come out.

$$\sqrt{A_1A_2} + \sqrt{B_1B_2} \leq \sqrt{A_1 + B_1}\sqrt{A_2 + B_2}$$

• What are $A$ and $B$? – Don Larynx May 24 '16 at 7:25
• Can you only use AM-GM? Cauchy looks simpler. – S.C.B. May 24 '16 at 7:31

For any positive real numbers $A_1,A_2,B_1$ and $B_2$ AM-GM inequality implies \begin{align*} 2\sqrt{A_1A_2B_1B_2}&\le A_1B_2+A_2B_1 \end{align*} Then \begin{align*} A_1A_2+2\sqrt{A_1A_2B_1B_2}+B_1B_2&\le A_1A_2+A_1B_2+A_2B_1+B_1B_2\\ \left(\sqrt{A_1A_2}+\sqrt{B_1B_2}\right)^2&\le\left(A_1+B_1\right)\left(A_2+B_2\right) \end{align*} So $$\sqrt{A_1A_2}+\sqrt{B_1B_2}\le\sqrt{A_1+B_1}\sqrt{A_2+B_2}$$
You first square both sides and then apply $AM-GM$ inequaity to the quantities $A_1B_2$ and $A_2B_1$