\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{xz}{cx+az}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)
Ta có:\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{xz}{cx+az}\)
\(\Rightarrow\frac{xyz}{ayz+bxz}=\frac{xyz}{bxz+cxy}=\frac{xyz}{cyx+ayz}\)
\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)
\(\Rightarrow\left\{\begin{matrix}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}ayz=cxy\\bxz=cxy\\bxz=ayz\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}az=cx\\bz=cy\\bx=ay\end{matrix}\right.\)\(\left(2\right)\)
Thay (2) vào (1) ta có:
\(\frac{xy}{2ay}=\frac{yz}{2bz}=\frac{xz}{2cx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right)\)
\(\Rightarrow\frac{x^2}{4a^2}=\frac{y^2}{4b^2}=\frac{z^2}{4c^2}=\frac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\frac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)
\(\Rightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{1}{4}\left(4\right)\). Thay (3) vào (2) ta có:
\(\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{1}{4}\)\(\Rightarrow\left\{\begin{matrix}x=\frac{a}{2}\\y=\frac{b}{2}\\z=\frac{c}{2}\end{matrix}\right.\)