Bài 2 :
Ta có x , y , z là các số thực dương
Khi đó : \(5\left(x^2+y^2+z^2\right)-9x\left(y+z\right)-18yz=0\)
\(\Leftrightarrow5\frac{x^2}{\left(y+z\right)^2}+\frac{5\left(y^2+z^2\right)}{\left(y+z\right)^2}-\frac{9x}{y+z}-\frac{18yz}{\left(y+z\right)^2}=0\)
\(\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-\frac{9x}{y+z}=\frac{18yz}{\left(y+z\right)^2}-\frac{5\left(y^2+z^2\right)}{\left(y+z\right)^2}\)
\(\le\frac{\frac{18\left(y+z\right)^2}{4}}{\left(y+z\right)^2}-\frac{\frac{5\left(y+z\right)^2}{2}}{\left(y+z\right)^2}=\frac{18}{4}-\frac{5}{2}=2\)
\(\Rightarrow5\left(\frac{x}{y+z}\right)^2-9.\frac{x}{y+z}\le2\)
Đặt \(\frac{x}{y+z}=a>0\) ta được : \(5a^2-9a-2\le0\)
\(\Leftrightarrow5a^2-10a+a-2\le0\Leftrightarrow\left(5a+1\right)\left(a-2\right)\le0\)
Dễ thấy :
\(5a+1>0\Rightarrow a-2\le0\Leftrightarrow a\le2\Leftrightarrow\frac{x}{y+z}\le2\)
Ta có :
\(Q=\frac{2x-y-z}{y+z}=\frac{2x}{y+z}-1\le2.2-1=3\)
Dấu " = '' xảy ra khi \(\left\{{}\begin{matrix}y=z\\\frac{x}{y+z}=2\end{matrix}\right.\) \(\Leftrightarrow x=4y=4z\)
Vậy GTLN của \(Q=3\Leftrightarrow x=4y=4z\)