Question

Cho x, y là các số thực dương thỏa mãn x+y <=3.

Tìm GTNN của \(A=\frac{2}{3xy}+\sqrt{\frac{3}{y+1}}\)

Answer 1

\(A=\frac{2}{3xy}+\sqrt{\frac{3}{y+1}}\)

\(A\ge\frac{2}{3xy}+\frac{1+\frac{3}{y+1}}{2}\left(AM-GM\right)\)

\(A\ge\frac{2}{3xy}+\frac{3}{2\left(y+1\right)}+\frac{1}{2}\)

Ta có:\(\frac{2}{3xy}+\frac{x}{3}+\frac{y}{6}\ge1\left(AM-GM\right)\)

\(\frac{3}{2\left(y+1\right)}+\frac{y+1}{6}\ge1\left(AM-GM\right)\)

Cộng vế theo vế \(\Rightarrow A\ge2-\frac{x}{3}-\frac{y}{6}-\frac{y+1}{6}+\frac{1}{2}\)

\(A\ge\frac{5}{2}-\frac{x+y}{3}-\frac{1}{6}\)

\(A\ge\frac{5}{2}-1-\frac{1}{6}=\frac{4}{3}\)

"="<=>\(x=1;y=2\)