Lower bound over a convex function I will be to grateful if help me find a tight lower bound $g(x)$ over the following convex function:
 $$f(x) = \sqrt{1+4x^2} -1 + \log(\sqrt{1+4x^2}-1) - \log(2x^2) \geq g(x),$$
where  $g(x)$ is preferably a polynomial with degree of at least $2$.
The taylor expansion around the point ($x = 0$) of this function is given by:
$$f(x) = x^2 - \frac{x^4}{2}  + \frac{2x^6}{3} - \frac{5x^8}{4} + \frac{14x^{10}}{5} - \ldots$$

For the case $|x|\leq 1$ I propose the following: Write $g(x):= a x^2- b x^4$ and choose $a$ and $b$ such that $f(1)=g(1)$, $f'(1)=g'(1)$. I did this using Mathematica and obtained
$$a=2\Bigl({3\over 1+\sqrt{5}} +\log {2\over 1+\sqrt{5}}\Bigr)\ \qquad b=
{2\over 1+\sqrt{5}} +\log{2\over 1+\sqrt{5}}\ .$$
Plotting  the resulting graph of $f-g$ one gets the impression that
$$0\leq f(x)-g(x)\leq 0.0117 \qquad(0\leq x\leq 1)\ .$$
