Question

x , y > 0 ; x + 2y\(\ge\)2

Tìm min A = 2x² + 16y² + \(\frac{2}{x}+\frac{3}{y}\)

Answer 1

Có đúng đề không ạ, \(2x^2\) hay \(2x\) ạ?

Answer 2

\(A=x^2+4y^2+x^2+\frac{1}{x}+\frac{1}{x}+12y^2+\frac{3}{2y}+\frac{3}{2y}\)

\(A\ge\frac{\left(x+2y\right)^2}{2}+3\sqrt[3]{\frac{x^2}{x^2}}+3\sqrt[3]{\frac{12y^2.3.3}{2y.2y}}\ge14\)

\(\Rightarrow A_{min}=14\) khi \(\left\{{}\begin{matrix}x=1\\y=\frac{1}{2}\end{matrix}\right.\)

Answer 3

\(A=2x^2+16y^2+\frac{2}{x}+\frac{3}{y}\\ =\left(x^2+\frac{1}{x}+\frac{1}{x}\right)+\left(12y^2+\frac{3}{2y}+\frac{3}{2y}\right)+\left(x^2+4y^2\right)\\ \overset{AM-GM}{\ge}3\sqrt[3]{x^2\cdot\frac{1}{x}\cdot\frac{1}{x}}+3\sqrt[3]{12y^2\cdot\frac{3}{2y}\cdot\frac{3}{2y}}+\frac{\left(x+2y\right)^2}{2}\\ \ge3+3\cdot3+\frac{2^2}{2}=14\)

Vậy............