Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=a^2k^2\\y^2=b^2k^2\\z^2=c^2k^2\end{matrix}\right.\)
Ta có: \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)\)
\(=\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)\)
\(=\left(a^2+b^2+c^2\right)^2\cdot k^2\)(1)
Ta có: \(\left(ax+by+cz\right)^2\)
\(=\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2\)
\(=\left(a^2k+b^2k+c^2k\right)^2\)
\(=\left(a^2+b^2+c^2\right)^2\cdot k^2\)(2)
Từ (1) và (2) suy ra \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)(đpcm)
