Ta có: \(N=\frac{a}{b+1}+\frac{b}{a+1}=\frac{a^2}{ab+a}+\frac{b^2}{ab+b}\)
\(\ge\frac{\left(a+b\right)^2}{a+b+2ab}\ge\frac{1}{1+\frac{\left(a+b\right)^2}{2}}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\)
Dấu = xảy ra khi \(a=b=\frac{1}{2}\)
Lại có: \(\frac{a}{b+1}=\frac{a}{2-a}\)
Do \(a;b\ge0\); a+b=1
\(\Rightarrow0\le a\le1\)
\(\Rightarrow2-a\ge1\)
\(\Rightarrow\frac{a}{2-a}\le a\left(a\ge0\right)\)
Tương tự suy ra \(N\le a+b=1\)
Dấu = xảy ra khi \(\left(a;b\right)=\left(0;1\right);\left(1;0\right)\)
Vậy \(N_{Min}=\frac{2}{3}\Leftrightarrow a=b=\frac{1}{2}\)
\(N_{Max}=1\Leftrightarrow\left(a;b\right)=\left(0;1\right);\left(1;0\right)\)
