Áp dụng BĐT Cauchy ta có:
$\dfrac{x+y+z}{2x}=\dfrac{1}{2}\left(\dfrac{y+z}{x}+1\right)\ge\sqrt{\dfrac{y+z}{x}}$
$\Rightarrow \sqrt{\dfrac{x}{y+z}}\ge\dfrac{2x}{x+y+z}$
Tương tự: $\sqrt{\dfrac{y}{x+z}}\ge\dfrac{2y}{x+y+z};\sqrt{\dfrac{z}{x+y}}\ge\dfrac{2z}{x+y+z}$
Từ đó suy ra:
$\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{x+z}}+\sqrt{\dfrac{z}{x+y}}\ge2>1$