Ta có:
$\dfrac{x^2}{x+yz}=\dfrac{x^3}{x^2+xy+yz+zx}=\dfrac{x^3}{(x+y)(x+z)}$
Áp dụng BĐT Cauchy ta có:
$\dfrac{x^3}{(x+y)(x+z)}+\dfrac{x+y}{8}+\dfrac{x+z}{8}\ge\dfrac{3x}{4} \Leftrightarrow \dfrac{x^3}{(x+y)(x+z)}\ge\dfrac{x}{2}-\dfrac{y+z}{8}$
Hay suy ra: $\dfrac{x^2}{x+yz}\ge\dfrac{x}{2}-\dfrac{y+z}{8}$
Tương tự: $\dfrac{y^2}{y+xz}\ge\dfrac{y}{2}-\dfrac{x+z}{8},\dfrac{z^2}{z+xy}\ge\dfrac{z}{2}-\dfrac{x+y}{8}$
Cộng 3 BĐT trên lại, ta có đpcm.
Dấu bằng xảy ra khi: $x=y=z=3$