Question

[Hoàng Diệu Anh]

áp dụng bdt cosi tìm gtnn của y=x/3+5/2x-1; x>1/2

Answer 1

[Dương Ngọc Nguyễn]

Answer 2

[Nguyễn Việt Lâm]

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

\(\Rightarrow y\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)

\(\Rightarrow P_{min}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)

Dấu "=" xảy ra khi \(\left(2x-1\right)^2=30\Rightarrow x=\frac{\sqrt{30}+1}{2}\)