\(\left(x+y-1\right)^2=xy\le\dfrac{1}{4}\left(x+y\right)^2\)
\(\Rightarrow\dfrac{3}{4}\left(x+y\right)^2-2\left(x+y\right)+1\le0\)
\(\Rightarrow\dfrac{2}{3}\le x+y\le2\)
Ta có:
\(P=\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{\sqrt{xy}}{x+y}\)
\(P\ge\dfrac{4}{2xy+x^2+y^2}+2\sqrt{\dfrac{\sqrt{xy}}{2xy\left(x+y\right)}}=\dfrac{4}{\left(x+y\right)^2}+2\sqrt{\dfrac{1}{2\sqrt{xy}\left(x+y\right)}}\)
\(P\ge\dfrac{4}{\left(x+y\right)^2}+2\sqrt{\dfrac{1}{\left(x+y\right)\left(x+y\right)}}=\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{x+y}\ge\dfrac{4}{2^2}+\dfrac{2}{2}=2\)
\(P_{min}=2\) khi \(x=y=1\)
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