Áp dụng Cauchy-Schwarz thì P\geq \frac{(x+y)^2}{2-(x+y)}+\frac{1}{x+y}+x+y=\frac{t^2}{2-t}+\frac{1}{t}+t=\frac{t^2}{2-t}+(\frac{1}{t}+\frac{9t}{4})-\frac{5t}{4} Với $00$
\Rightarrow P\geq \frac{4t^2-5t(2-t)}{4(2-t)}+2.\sqrt{\frac{9t}{4t}}=\frac{9t^2-10t}{4(2-t)}+3
Ta cm \frac{9t^2-10t}{4(2-t)}\geq \frac{-1}{2} (*)
Thật vậy (*)\Leftrightarrow \frac{9(t-\frac{2}{3})^2}{2(2-t)}\geq 0 đúng do 2-t>0
\Rightarrow P\geq \frac{5}{2}
Dấu "=" xảy ra khi x=y=\frac{1}{3}
Vậy Min P=\frac{5}{2}