Hi everyone, I will be to grateful if help me find a tight lower bound $g(x)$ over the following concave function: $$f(x) = \sqrt{1+4x} 1 + \log(\sqrt{1+4x}1)  \log(2x) \geq g(x),$$ where $x \geq 0$. The taylor expansion around the point ($x = 0$) of this function is given by: $$f(x) = x  \frac{x^2}{2}  + \frac{2x^3}{3}  \frac{5x^4}{4} + \frac{14x^{5}}{5}  \ldots$$
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