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$
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$