Question

[l҉o҉n҉g҉ d҉z҉]

Cho x , y , z > 0 . Chứng minh \(\frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\ge x+y+z\)

Answer 1

[Tran Le Khanh Linh]

Áp dùng BĐT Cosi ta có:

\(\frac{x^3}{yz}+y+z\ge3\sqrt[3]{\frac{x^3}{yz}\cdot y\cdot z}=3x\)

\(\frac{y^3}{xz}+z+x\ge3\sqrt[3]{\frac{z^3}{zx}\cdot z\cdot x}=3y\)

\(\frac{z^3}{yx}+x+y\ge3\sqrt[3]{\frac{z^3}{xy}\cdot x\cdot y}=3z\)

\(\Rightarrow\frac{x^3}{xy}+y+z+\frac{y^3}{zx}+x+z+\frac{z^3}{xy}+x+y\ge3x+3y+3z\)

\(\Rightarrow\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\ge3\left(x+y+z\right)-2\left(x+y+z\right)\)\(=x+y+z\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^3}{yz}=y=z\\\frac{y^3}{zx}=x=z\\\frac{z^3}{yz}=y=x\end{cases}\Rightarrow x=y=z}\)