Question

Cho x, y là các sô thực dương thỏa mãn \(x+y\le3\). Tìm GTNN của biểu thức:

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

Answer 1

\(x+y\le3\Rightarrow x\le3-y\Rightarrow\frac{1}{x}\ge\frac{1}{3-y}\)

\(\Rightarrow A\ge\frac{2}{3y\left(3-y\right)}+\frac{6}{2\sqrt{3\left(y+1\right)}}\ge\frac{2}{3y\left(3-y\right)}+\frac{6}{y+4}\)

\(A\ge\frac{2}{3y\left(3-y\right)}+\frac{6}{y+4}-\frac{4}{3}+\frac{4}{3}\)

\(A\ge\frac{2\left(2y^3-7y^2+4y+4\right)}{3y\left(3-y\right)\left(y+4\right)}+\frac{4}{3}=\frac{2\left(y-2\right)^2\left(2y+1\right)}{3y\left(3-y\right)\left(y+4\right)}+\frac{4}{3}\ge\frac{4}{3}\)

\(A_{min}=\frac{4}{3}\) khi \(\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)