Question

Với \(x>y>0\), tìm giá trị nhỏ nhất:

\(N=2x+\frac{32}{\left(x-y\right)\left(2y+3\right)^2}\)

Answer 1

\(N=2\left(x-y\right)+\frac{32}{\left(x-y\right)\left(2y+3\right)^2}+2y\)

\(\Rightarrow N\ge2\sqrt{2\left(x-y\right)\frac{32}{\left(x-y\right)\left(2y+3\right)^2}}+2y\)

\(\Rightarrow N\ge\frac{16}{2y+3}+2y=\frac{16}{2y+3}+2y+3-3\)

\(\Rightarrow N\ge2\sqrt{\frac{16}{\left(2y+3\right)}.\left(2y+3\right)}-3=8-3=5\)

\(\Rightarrow N_{min}=5\) khi \(\left\{{}\begin{matrix}x=\frac{3}{2}\\y=\frac{1}{2}\end{matrix}\right.\)