Question

Cho x, y , z \(\ne\) 0 và \(x^2=yz\),\(y^2=xz\),\(z^2=xy\). Chứng minh x = y = z.

Answer 1

Ta có: \(\left\{{}\begin{matrix}x^2=yz\\y^2=xz\\z^2=xy\end{matrix}\right.\)

Cộng theo vế 3 đẳng thức trên ta có:

\(x^2+y^2+z^2=yz+xz+xy\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2yz+2xz+2xy\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2yz-2xz-2xy=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)\(\Rightarrow x=y=z\)