\(\dfrac{a+b}{2}\ge\sqrt{ab}\Rightarrow a+b\ge2\sqrt{ab}\)
\(\Rightarrow\left(a+b\right)^2\ge\left(2\sqrt{ab}\right)^2\)
\(\Rightarrow a^2+2ab+b^2\ge4ab\)
\(\Rightarrow a^2-2ab+b^2\ge0\Rightarrow\left(a-b\right)^2\ge0\forall x\)
Đẳng thức xảy ra khi \(a=b\)
Ta có: \(\left(\sqrt{a}+\sqrt{b}\right)\ge0\) với mọi a và b \(\left(a,b\ge0\right)\)
\(\Leftrightarrow a+2\sqrt{ab}+b\ge0\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)
\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\) (đpcm)
Vậy \(\dfrac{a+b}{2}\ge\sqrt{ab}\)
