Ta có :\(a+b+\dfrac{1}{2}=a+b+\dfrac{1}{4}+\dfrac{1}{4}=\left(a+\dfrac{1}{4}\right)+\left(b+\dfrac{1}{4}\right)\)
Áp dụng bất đẳng thức cô si ta có :
\(a+\dfrac{1}{4}\ge2\sqrt{a.\dfrac{1}{4}}=\sqrt{a}\)
\(b+\dfrac{1}{4}\ge2\sqrt{b.\dfrac{1}{4}}=\sqrt{b}\)
Do đó :\(a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)
Dấu "=" xảy ra khi :\(a=b=\dfrac{1}{4}\)
Vậy với \(a,b\ge0\) thì \(a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)
Ta có: \(a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)
\(\Leftrightarrow\left(a-2\sqrt{a}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\left(b-2\sqrt{b}.\dfrac{1}{2}+\dfrac{1}{4}\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\left(\sqrt{b}-\dfrac{1}{2}\right)^2\ge0\) ( luôn đúng )
\(\Rightarrowđpcm\)
