Áp dụng bất đẳng thức $Schwarz$ ta được:
$\dfrac{a + b}{2} = \dfrac{a}{2} + \dfrac{b}{2}$
$\geq \dfrac{(\sqrt a + \sqrt b)^2}{2+2} = \left(\dfrac{\sqrt a + \sqrt b}{2}\right)^2$
$\geq \left[\left(\dfrac{\sqrt[4]{a} + \sqrt[4]{b}}{2}\right)^2\right]^2 = \left(\dfrac{\sqrt[4]{a} + \sqrt[4]{b}}{2}\right)^4$
$\Rightarrow \sqrt[4]{\dfrac{a + b}{2}}\geq \dfrac{\sqrt[4]{a} + \sqrt[4]{b}}{2}$
Dấu = xảy ra $\Leftrightarrow a = b$
