\(x,y>0;x+y=2\)
\(P=\dfrac{x^2+y^2+1}{\left(2x^2+1\right)\left(2y^2+1\right)}+\dfrac{1}{xy}\)
\(=\dfrac{1}{2}.\dfrac{2x^2+2y^2+2}{\left(2x^2+1\right)\left(2y^2+1\right)}+\dfrac{1}{xy}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2x^2+1}+\dfrac{1}{2y^2+1}\right)+\dfrac{1}{xy}\)
\(\ge^{CBS}\dfrac{1}{2}.\dfrac{4}{2\left(x^2+y^2+1\right)}+\dfrac{1}{xy}\)
\(=\dfrac{1}{x^2+y^2+1}+\dfrac{1}{3xy}+\dfrac{2}{3xy}\)
\(\ge^{CBS}\dfrac{\left(1+1\right)^2}{\left(x+y\right)^2+xy+1}+\dfrac{2}{3xy}\)
\(=\dfrac{4}{xy+5}+\dfrac{2}{3xy}\ge^{Cauchy}\dfrac{4}{\dfrac{\left(x+y\right)^2}{4}+5}+\dfrac{2}{3.\dfrac{\left(x+y\right)^2}{4}}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(x=y=1\)
Vậy giá trị nhỏ nhất của P là 4/3.
