Question

xy+xz+5yz=17,tìm min P=X^2+33Y^2+33Z^2

Answer 1

\(x^2+36y^2\ge12xy\) ; \(x^2+36z^2\ge12zx\) ; \(30y^2+30z^2\ge60yz\)

Cộng vế với vế:

\(2x^2+66y^2+66z^2\ge12xy+12yz+60zx\)

\(\Leftrightarrow2\left(x^2+33y^2+33z^2\right)\ge12\left(xy+yz+5zx\right)=204\)

\(\Rightarrow P\ge102\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x^2=36y^2=36z^2\\xy+zx+5yz=17\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=z^2=1\end{matrix}\right.\)