Lower bound over a concave function

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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What do require exactly on the function $g$? –  Pietro Majer May 30 '11 at 17:59
To be quite tight, and preferably $g(x)=ax+b$ where $b \geq 1$. –  Farzad May 30 '11 at 18:37
Is $g(x)$ of interest near 0, for large values, or somewhere in between? For large $x$, the function is dominated by the $\sqrt{4x+1}$ term so a linear bound is going to be poor; you need to define the range in which the bound is of interest. If the linearity requirement is dropped, clearly $g(x) = 2\sqrt{x} - 1$ is going to be a fairly good lower bound for all large enough $x$. –  András Salamon Sep 23 '13 at 15:19
and $ax+b$ with $b\geq1$ is impossible near zero, from your taylor expansion. –  Athanagor Wurlitzer Jan 29 at 14:49