\(< =>\dfrac{30x^2}{60}+\dfrac{20y^2}{60}+\dfrac{15z^2}{60}=\dfrac{12x^2+12y^2+12z^2}{60}\\ < =>30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\\ < =>18x^2+8y^2+3z^2=0\)
có \(\left\{{}\begin{matrix}x^2\ge0\\y^2\ge0\\z^2\ge0\end{matrix}\right.< =>\left\{{}\begin{matrix}18x^2\ge0\\8y^2\ge0\\3z^2\ge0\end{matrix}\right.\)
`=>18x^2+8y^2+3z^2>=0`
dấu ''='' xảy khi \(\left\{{}\begin{matrix}18x^2=0\\8y^2=0\\3z^2=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)
\(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\\ \Rightarrow\dfrac{30x^2+20y^2+15z^2}{60}=\dfrac{12x^2+12y^2+12z^2}{60}\\ \Rightarrow18x^2+8y^2+3z^2=0\)
Do \(x^2\ge0\forall x;y^2\ge0\forall y;z^2\ge0\forall z\)
\(\Rightarrow18x^2+8y^2+3z^2\ge0\)
Dấu "=" xảy ra khi \(x=y=z=0\)
