Question

Mn giúp mình với ạ , cám ơn trước nhé .

Giả sử x, y, z là các số thực thỏa mãn hệ thức \(\left\{{}\begin{matrix}x^2+xz+z^2=3\\y^2+yz+z^2=16\end{matrix}\right.\)

Tìm S max biết S = xy+yz+xz

Answer 1

\(S^2=\left(xy+yz+zx\right)^2=\left[x\left(y+\frac{z}{2}\right)+\left(y+\frac{x}{2}\right)z\right]^2\)

\(S^2=\left[x\left(y+\frac{z}{2}\right)+\left(\frac{2}{\sqrt{3}}y+\frac{1}{\sqrt{3}}x\right).\left(\frac{\sqrt{3}}{2}z\right)\right]^2\)

\(S^2\le\left[x^2+\left(\frac{2}{\sqrt{3}}y+\frac{1}{\sqrt{3}}x\right)^2\right]\left[\left(y+\frac{z}{2}\right)^2+\frac{3}{4}z^2\right]\)

\(S^2\le\left(x^2+\frac{4}{3}y^2+\frac{4}{3}xy+\frac{1}{3}x^2\right)\left(y^2+yz+\frac{z^2}{4}+\frac{3}{4}z^2\right)\)

\(S^2\le\frac{4}{3}\left(x^2+xy+y^2\right)\left(y^2+yz+z^2\right)=64\)

\(\Rightarrow S\le8\Rightarrow S_{max}=8\)