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)=6dấu"="\Leftrightarrow x=y$9.Ta có (2√x+1+√x−2)2≤(22+12)(2x−1)=5(2x−1)BDTD≤(x+2)2⇒2√x+1+√x−2−x≤2⇔maxB=2016
6. P=(x+y)2(1x2+y2+1xy) =(x+y)2(1x2+y2+12xy+12xy) ≥(x+y)2(4(x+y)2+1(x+y)22=6 dấu "=" ⇔x=y
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√x+1+√x−2)2≤(22+12)(2x−1)=5(2x−1)BDTD≤(x+2)2⇒2√x+1+√x−2−x≤2⇔maxB=2016