\(DK:a,b\ge0\)
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\Leftrightarrow\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\ge\sqrt{a+b}^2\)
\(\Leftrightarrow a+2\sqrt{ab}+b\ge a+b\)
\(\Leftrightarrow2\sqrt{ab}\ge0\forall a,b\)
\(\Rightarrow\left|a\right|+\left|b\right|\ge\left|a+b\right|\)