Question

[Dương Thanh Ngân]

Tìm GTNN của biểu thức:

\(A=\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{x+z}\)

Biết\(\left\{{}\begin{matrix}x.y.z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\end{matrix}\right.\)

Answer 1

[Nguyễn Việt Lâm]

\(A\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{1}{2}\left(x+y+z\right)\ge\dfrac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\dfrac{1}{2}\)

\(A_{min}=\dfrac{1}{2}\) khi \(x=y=z=\dfrac{1}{3}\)