Làm
- Vì (x+1)(y+1) =2 \(\Leftrightarrow\) xy+x+y+1 =2
\(\Leftrightarrow\) xy+ (x+y) = 1
Đặt x+y = a, xy=b \(\Rightarrow\) a+b = 1 \(\Rightarrow\) 1-b \(=\) a
và a,b \(\ge0\) và x,y không âm
- Có: \(x^2+y^2-2\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+2\)
= \(\left(x+y\right)^2-2xy-\sqrt{2\left(x^2y^2+x^2+y^2+1\right)}+2\)
= \(\left(x+y\right)^2-2xy+2-\sqrt{2\left[x^2y^2+\left(x+y\right)^2-2xy+1\right]}\)
=\(a^2-2b+2-\sqrt{2\left(b^2+a^2-2b+1\right)}\)
= \(a^2-2b+2-\sqrt{2\left(1-b\right)^2+a^2}\)
=\(a^2+2\left(1-b\right)-\sqrt{2\left(1-b\right)^2+a^2}\)
= \(a^2+2a-\sqrt{2.2a^2}\)
= \(a^2+2a-2\left|a\right|\)
= \(a^2+2a-2a\)
= \(a^2\)
\(\Rightarrow P=\sqrt{a^2}+b=\left|a\right|+b=a+b=1\)
Kl: P =1
