Có:
\(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x=y+z\\-y=x+z\\-z=x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{matrix}\right.\)
\(\Rightarrow ax^2+by^2+cz^2\)
\(=a\left(y+z\right)^2+b\left(x+z\right)^2+c\left(x+y\right)^2\)
\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bxz+cxy\right)\)
Mà \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\a+c=-b\\a+b=-c\end{matrix}\right.\)
Đồng thời có: \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
Từ đây ta có:)
\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)
\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)
\(\Rightarrow ax^2+by^2+cz^2=0\left(đpcm\right)\)
