Ta có: \(\dfrac{x^2}{6}+\dfrac{y^2}{7}+\dfrac{z^2}{8}=\dfrac{x^2+y^2+z^2}{10}\)
\(\Rightarrow\dfrac{x^2}{6}+\dfrac{y^2}{7}+\dfrac{z^2}{8}=\dfrac{x^2}{10}+\dfrac{y^2}{10}+\dfrac{z^2}{10}\)
\(\Rightarrow\dfrac{x^2}{6}+\dfrac{y^2}{7}+\dfrac{z^2}{8}-\dfrac{x^2}{10}-\dfrac{y^2}{10}-\dfrac{z^2}{10}=0\)
\(\Rightarrow\left(\dfrac{5x^2}{30}-\dfrac{3x^2}{40}\right)+\left(\dfrac{10y^2}{70}-\dfrac{7y^2}{70}\right)+\left(\dfrac{5z^2}{40}-\dfrac{4z^2}{40}\right)=0\)
\(\Rightarrow\dfrac{x^2}{15}+\dfrac{3y^2}{70}+\dfrac{z^2}{40}=0\)
Mà x2 , y2 , z2 \(\ge\) 0 \(\Rightarrow\dfrac{x^2}{15}+\dfrac{3y^2}{70}+\dfrac{z^2}{40}\ge0\)
Dấu "=" xảy ra khi x = y = z = 0
Vậy \(x=y=z=0\Rightarrow\dfrac{x^2}{6}+\dfrac{y^2}{7}+\dfrac{z^2}{8}=\dfrac{x^2+y^2+z^2}{10}\)
