Question

[hung]

Cho \(a,b\ge0\)thỏa mãn a+b=2   Tìm GTNN,GTLN của

\(M=a\sqrt{b+1}+b\sqrt{a+1}\)

Answer 1

[pham trung thanh]

Ta có: \(a\sqrt{b+1}=\frac{a\sqrt{\left(b+1\right)2}}{\sqrt{2}}\le a\frac{b+1+2}{2\sqrt{2}}=\frac{ab+3a}{2\sqrt{2}}\)

Tương tự: \(b\sqrt{a+1}\le\frac{ab+3b}{2\sqrt{2}}\)

\(\Rightarrow M\le\frac{3\left(a+b\right)+2ab}{2\sqrt{2}}\le\frac{6+\frac{\left(a+b\right)^2}{2}}{2\sqrt{2}}=\frac{8}{2\sqrt{2}}=2\sqrt{2}\)

Dấu = khi a=b=1

Ta có: \(a+b=2\Rightarrow b=2-a\)

\(\Rightarrow a\sqrt{b+1}=a\sqrt{3-a}\)

Lại có: \(\hept{\begin{cases}a;b>0\\a+b=2\end{cases}}\Rightarrow0\le a;b\le2\)

Mặt khác: \(a\le2\Rightarrow3-a\ge1\)

\(\Rightarrow\sqrt{3-a}\ge1\)

\(\Rightarrow a\sqrt{3-a}\ge a\)     Do \(a\ge0\)

Tương tự suy ra \(M\ge a+b=2\)

Dấu = khi \(\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)

Vậy \(M_{Max}=2\sqrt{2}\Leftrightarrow a=b=1\)

        \(M_{Min}=2\Leftrightarrow\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)