6. P=(x+y)2(1x2+y2+1xy) $ =(x+y)^{2}\left(\frac{1}{x^{2}+y^{2}}+\frac{1}{2xy}+\frac{1}{2xy}\right)\geq (x+y)^{2}\left(\frac{4}{(x+y)^{2}}+\frac{1}{\frac{(x+y)^{2}}{2}}\right)=6 dấu "=" \Leftrightarrow x=y$9.Ta có (2\sqrt{x+1}+\sqrt{x-2})^2 \le (2^2+1^2)(2x-1)=5(2x-1) \overset{BDTD}{\le} (x+2)^2\Rightarrow 2\sqrt{x+1}+\sqrt{x-2} -x\le 2\Leftrightarrow \max B=2016
6. P=(x+y)^{2}(\frac{1}{x^{2}+y^{2}}+\frac{1}{xy}) =(x+y)^{2}(\frac{1}{x^{2}+y^{2}}+\frac{1}{2xy}+\frac{1}{2xy}) \geq (x+y)^{2}(\frac{4}{(x+y)^{2}}+\frac{1}{\frac{(x+y)^{2}}{2}}=6 dấu "=" \Leftrightarrow x=y
6.
P=(x+y)^{2}(\frac{1}{x^{2}+y^{2}}+\frac{1}{xy}) $ =(x+y)^{2}
\left(\frac{1}{x^{2}+y^{2}}+\frac{1}{2xy}+\frac{1}{2xy}
\right)
\geq (x+y)^{2}
\left(\frac{4}{(x+y)^{2}}+\frac{1}{\frac{(x+y)^{2}}{2}}
\right)=6
dấu "=" \Leftrightarrow x=y$
9.Ta có (2\sqrt{x+1}+\sqrt{x-2})^2 \le (2^2+1^2)(2x-1)=5(2x-1) \overset{BDTD}{\le} (x+2)^2\Rightarrow 2\sqrt{x+1}+\sqrt{x-2} -x\le 2\Leftrightarrow \max B=2016