a) Ta có:
\(\left(\sqrt{a}+\sqrt{b}\right)^2=\left(\sqrt{a}\right)^2+2\sqrt{a}.\sqrt{b}+\left(\sqrt{b}\right)^2=a+2\sqrt{a}.\sqrt{b}+b\)
\(\left(\sqrt{a+b}\right)^2=a+b\)
Vì \(a+2\sqrt{a}.\sqrt{b}+b>a+b\) nên \(\left(\sqrt{a}+\sqrt{b}\right)^2>\left(\sqrt{a+b}\right)^2\). \(\Rightarrow\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)