If α ∈ (0, π/2), then minimum value of \(2x+\frac{tan^2α}{2x}\) is

Concept:

AM, GM, HM Formulas:

If A is the arithmetic mean 

⇔ \({\rm{A}} = \frac{{{\rm{a\;}} + {\rm{\;b}}}}{2}\)

If G is the geometric mean 

⇔ \({\rm{G}} = \sqrt {{\rm{ab}}} \)

Relation between AM, and GM

AM  ≥  GM

Calculation:

Given that,

\(2x+\frac{tan^2α}{2x}\)

Let a = 2x, b = \(\frac{tan^2α}{2x}\)

We know that,

AM  ≥  GM

\(\frac{2x+\frac{tan^2α}{2x}}{2} ≥ \ \sqrt{2x\times\frac{tan^2α}{2x}}\)

⇒ \(2x+\frac{tan^2α}{2x}\) ≥ 2 tan α 

⇒  \(2x+\frac{tan^2α}{2x}\) ∈ [2 tan α, ∞)

Hence, the minimum value is 2 tan α​.