TITLE: Does $\sqrt{a+b} \le \sqrt a + \sqrt b$ hold for all positive real numbers a and b?
QUESTION [15 upvotes]: I thought of this a while ago, but can't make up a proof or a counterexample. Does anyone know more about this?
$$\sqrt{a+b}  \le \sqrt a + \sqrt b , \forall a,b \in \mathbb R_+$$
Moreover, what happens with more variables? Say:
$$\sqrt{x_1+x_2+...+x_n} \le \sqrt x_1 +\sqrt x_2 + ... + \sqrt x_n $$
with $ x_i \in\mathbb R_+ \forall i \in \{1,2,...,n\} $
Or when $i = \mathbb N$
PS: I tried looking on the internet for this but I don't know how I would call this.

REPLY [5 votes]: Hint
$$\sqrt{a+b} \leq \sqrt{a+2\sqrt{ab}+b}=\sqrt{(\sqrt{a}+\sqrt{b})^2}$$
Same way
$$\sqrt{x_1+...+x_n} \leq \sqrt{x_1+...+x_n+2(\sqrt{x_1x_2}+\sqrt{x_1x_2}+..+\sqrt{x_{n-1}x_n})}=\sqrt{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}$$

REPLY [2 votes]: Suppose you have proved the inequality for $n=2$. Suppose it holds for $n\ge2$; then
$$\def\vA{\vphantom{A}}
\sqrt{x_1+\dots+x_n+x_{n+1}\vA}
\le
\sqrt{x_1+\dots+x_n\vA}+\sqrt{x_{n+1}\vA}
\le
\sqrt{x_1\vA}+\dots+\sqrt{x_n\vA}+\sqrt{x_{n+1}\vA}
$$
The first $\le$ is from the case $n=2$, the second one from the induction hypothesis.
For the $n=2$ case, just square.