\(\left(\sqrt{A}+\sqrt{B}\right)^2=A+B+2\sqrt{AB}\)
\(\left(\sqrt{A+B}\right)^2=A+B\)
mà \(2\sqrt{AB}>0\)
nên \(\sqrt{A}+\sqrt{B}>\sqrt{A+B}\)
\(\left(\sqrt{A}+\sqrt{B}\right)^2=A+B+2\sqrt{AB}\\ \left(\sqrt{A+B}\right)^2=A+B\\ 2\sqrt{AB}\ge0\Leftrightarrow A+B+2\sqrt{AB}\ge A+B\\ \Leftrightarrow\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\)