Sum of square roots

Cauchy-Schwarz.
$$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2\le(1^2+1^2+1^2)(a+b+c)=3(a+b+c).$$

Conclusion. Taking square roots gives the result, with equality iff $a=b=c$.

General form. For $n$ terms, $\sum\sqrt{a_i}\le\sqrt{n\sum a_i}$.