How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$ How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$.
Can this be proved with simple way?

$$\left(\frac{\sin x}{x}\right)^2 + \left(\frac{\tan x}{x}\right) = \\ \left(\frac{x - \frac{x^3}{6} + \frac{x^5}{120} + \ldots}{x}\right)^2 + \left(\frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots}{x}\right) = \\ \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \ldots\right)^2 + \left(1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\right) \simeq \\ \left(1 + 2 \left(-\frac{x^2}{6} + \frac{x^4}{120} - \ldots \right)\right) + 1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\\2 + x^2 \left(\frac{ - 2}{6} + \frac{1}{3}\right) + x^4 \left(\frac{2}{120} + \frac{2}{15} \right) + \ldots = \\2 + x^4 \left(\frac{9}{60} \right) + \ldots > 2.$$
