Question

B= x + \(\dfrac{1}{x}\)với x > hoặc bằng 2

 

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Answer 1

Ta có:\(B=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3x}{4}\)

Áp dụng bất đẳng thức Cô-si ta có:

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

Ta có: \(\dfrac{3x}{4}\ge\dfrac{3.2}{4}=\dfrac{3}{2}\)

\(\Rightarrow B=1+\dfrac{3}{2}=\dfrac{5}{2}\)

Dấu "=" xảy ra ⇔ x=2

Vậy \(MinB=\dfrac{5}{2}\Leftrightarrow x=2\)

Answer 2

\(B=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3}{4}x\ge2\sqrt{\dfrac{x}{4x}}+\dfrac{3}{4}.2=1+\dfrac{3}{2}=\dfrac{5}{2}\)(do \(x\ge2\))

\(minB=\dfrac{5}{2}\Leftrightarrow x=2\)