\(a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)
\(\Rightarrow a+b+\dfrac{1}{4}+\dfrac{1}{4}-\sqrt{a}-\sqrt{b}\ge0\)
\(\Rightarrow\left(a-\sqrt{a}+\dfrac{1}{4}\right)+\left(b-\sqrt{b}+\dfrac{1}{4}\right)\ge0\)
\(\Rightarrow\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\left(\sqrt{b}-\dfrac{1}{2}\right)^2\ge0\) (đúng)
\("="\Leftrightarrow a=b=\dfrac{1}{4}\)