Question

[Lương Thế Vinh]

Cho x > 0. Tìm GTNN của C = x + 1/(4x) + x/((2x+1)^2)

Answer 1

[Trần Tuấn Hoàng]

\(x>0\)

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

-Ta đặt \(A=T=4x^2+1;B=4x\) thì ta có: 

\(A\ge B\Rightarrow A+T\ge B+T\) (do \(T>0\))\(\Rightarrow\dfrac{A+T}{B+T}\ge1\)

-Do đó: \(C=\dfrac{4x^2+1}{4x}+\dfrac{x}{\left(2x+1\right)^2}\ge\text{​​​​}\dfrac{4x^2+1+4x^2+1}{4x+4x^2+1}+\dfrac{x}{\left(2x+1\right)^2}=\dfrac{2\left(4x^2+1\right)}{\left(2x+1\right)^2}+\dfrac{8x}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=\dfrac{2\left(2x+1\right)^2}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=2-\dfrac{7x}{\left(2x+1\right)^2}\)

-Áp dụng BĐT AM-GM ta có:

\(C\ge2-\dfrac{7x}{\left(2x+1\right)^2}\ge2-\dfrac{7x}{4.2x}=2-\dfrac{7}{8}=\dfrac{9}{8}\)

\(C=\dfrac{9}{8}\Leftrightarrow x=\dfrac{1}{2}\)

-Vậy \(C_{min}=\dfrac{9}{8}\)