Question

Biết `x,y,z>0`, \(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\). CMR: \(x=y=z\)

Answer 1

\(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow2x+2y+2z=2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}\)

\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{yz}+z\right)+\left(z-2\sqrt{zx}+x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=\sqrt{y}\\\sqrt{y}=\sqrt{z}\\\sqrt{z}=\sqrt{x}\end{matrix}\right.\)

\(\Rightarrow x=y=z\)