TITLE: Proof of $\sqrt{a} + \sqrt{b} \le 2 \times \sqrt{a+b}$?
QUESTION [5 upvotes]: I want to prove that  $\sqrt{a} + \sqrt{b} \le 2 \times \sqrt{a+b}$, I had the idea to draw it:

Would it be enough to prove what I want to prove? If not, is there a way to be more precise by still using my method or should I abandon it and use a more "traditional" way?  
Thank you.

REPLY [2 votes]: Not all problems have a nice geometrical interpretation. I think it is surely better to do this one in a purely algebraic way.
It would also be more convenient to assume a,b $\geq$ 0 for your problem.
Because a,b$\geq$0, then a$\leq$a+b and b$\leq$a+b, and because f(x)=$\sqrt{x}$ is increasing on $ R _ +$, then $\sqrt{a} \leq \sqrt{a+b}$ and $\sqrt{b} \leq \sqrt{a+b}.$
Finally, by adding the 2 inequalities, we get $\sqrt{a}+\sqrt{b} \leq 2*\sqrt{a+b}$