Question

[hung]

Cho \(a,b\ge0\)thỏa mãn a+b=1  Tìm GTLN,GTNN của 

\(N=\frac{a}{b+1}+\frac{b}{a+1}\)

Answer 1

[pham trung thanh]

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 \(N_{Min}=\frac{2}{3}\Leftrightarrow a=b=\frac{1}{2}\)

    \(N_{Max}=1\Leftrightarrow\left(a;b\right)=\left(0;1\right);\left(1;0\right)\)