ta có: $x+y=1\Rightarrow x=1-y;y=1-x$ thay vào $P$$P=\frac{1-y}{y+1}+\frac{1-x}{x+1}=2(\frac{1}{x+1}+\frac{1}{y+1})-2\geq \frac{8}{x+y+2}-2=\frac{2}{3}$"=" khi $x=y=\frac{1}{2}$
ta có: $x+y=1\Rightarrow x=1-y;y=1-x$ thay vào $P$$P=\frac{1-y}{y+1}+\frac{1-x}{x+1}=2(\frac{1}{x+1}+\frac{1}{y+1})-2\geq \frac{4}{x+y+2}-2=\frac{2}{3}$"=" khi $x=y=\frac{1}{2}$
ta có: $x+y=1\Rightarrow x=1-y;y=1-x$ thay vào $P$$P=\frac{1-y}{y+1}+\frac{1-x}{x+1}=2(\frac{1}{x+1}+\frac{1}{y+1})-2\geq \frac{
8}{x+y+2}-2=\frac{2}{3}$"=" khi $x=y=\frac{1}{2}$