- Áp dụng BĐT Caushy ta có:
\(\dfrac{x^3}{y\left(z+x\right)}+\dfrac{y}{2}+\dfrac{z+x}{4}\ge\dfrac{3x}{2}\left(1\right)\)
- Tương tự: \(\left\{{}\begin{matrix}\dfrac{y^3}{z\left(x+y\right)}+\dfrac{z}{2}+\dfrac{x+y}{4}\ge\dfrac{3y}{2}\left(2\right)\\\dfrac{z^3}{x\left(y+z\right)}+\dfrac{x}{2}+\dfrac{y+z}{4}\ge\dfrac{3z}{2}\left(3\right)\end{matrix}\right.\)
- Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\) ta có:
\(\dfrac{x^3}{y\left(z+x\right)}+\dfrac{y^3}{z\left(x+y\right)}+\dfrac{z^3}{x\left(y+z\right)}+x+y+z\ge\dfrac{3\left(x+y+z\right)}{2}\)
\(\Leftrightarrow\dfrac{x^3}{y\left(z+x\right)}+\dfrac{y^3}{z\left(x+y\right)}+\dfrac{z^3}{x\left(y+z\right)}\ge\dfrac{x+y+z}{2}\)
- Mà theo BĐT Caushy ta có:
\(x+y+z\ge3\sqrt[3]{xyz}=3\sqrt[3]{1}=3\)
\(\Rightarrow\dfrac{x^3}{y\left(z+x\right)}+\dfrac{y^3}{z\left(x+y\right)}+\dfrac{z^3}{x\left(y+z\right)}\ge\dfrac{3}{2}\left(đpcm\right)\)
- Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
