Question

Tìm GTNN của: C= x + \(\dfrac{1}{4x}\)+ \(\dfrac{x}{\left(2x+1\right)^{ }2}\)

Answer 1

P/s : Mik nghĩ là \(\left(2x+1\right)^2\)

\(C=x+\dfrac{1}{4x}+\dfrac{x}{\left(2x+1\right)^2}=\left[\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\right]+\dfrac{3}{4}\left(x+\dfrac{1}{4x}\right)-\dfrac{1}{8}\)

AD BĐT AM - GM ta được : \(\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\ge4\sqrt[4]{\dfrac{1}{16^3}}=\dfrac{1}{2}\)

\(x+\dfrac{1}{4x}\ge2\sqrt{\dfrac{1}{4}}=1\) 

Suy ra : \(C\ge\dfrac{1}{2}+\dfrac{3}{4}.1-\dfrac{1}{8}=\dfrac{9}{8}\)

" = " \(\Leftrightarrow x=\dfrac{1}{2}\)