Does $\sqrt{a+b} \le \sqrt a + \sqrt b$ hold for all positive real numbers a and b?

Hint:$$\big(\sqrt{a+b}\big)^2 - \big(\sqrt{a}+\sqrt{b}\big)^2 = \big(a+b\big)-\big(a+b+2\sqrt{ab}\big)=-2\sqrt{ab}$$

Hint $$\sqrt{a+b} \leq \sqrt{a+2\sqrt{ab}+b}=\sqrt{(\sqrt{a}+\sqrt{b})^2}$$

Same way $$\sqrt{x_1+...+x_n} \leq \sqrt{x_1+...+x_n+2(\sqrt{x_1x_2}+\sqrt{x_1x_2}+..+\sqrt{x_{n-1}x_n})}=\sqrt{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}$$

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.