I am working with a log-concave real random variable, that has a density $f(x) = \exp(-\varphi(x))$ with $\varphi$ convex. Assuming that $X$ is centered and has unit variance ($\mathbb{E}X=0$, $\mathbb{E}X^2=1$), I want to prove that $X$ satisfies a **small deviation** inequality like
$$\mathbb{P}(|X| \leq t) \geq c t$$
for a numerical constant $c$ and **all $t\in (0,1)$** (emphasize on $t\leq 1$).
I know that log-concave distributions have sub exponential tails but this allows to prove large deviation bounds, typically $\mathbb{P}(|X| > t)\leq 2\exp(-ct)$.

My progress so far:

the density satisfies $f(0)\geq 1/8$ (see Lovasz and Vempala, lemma 5.5) so I expect that some regularity of the density $f$ (or the potential $\varphi$) should allow to prove that $f$ is uniformly

**lower-bounded**on the interval $[-1, 1]$.Essentially, if $f$ decays too fast around $0$, it would cause the variance to be too small, and we set the variance to $1$.I know how to prove the result if $\varphi$ (hence $f$) is even, because in this case we can use the lower bound for the concentration function of Bobkov and Chistyakov, corollary 2.2. Is it possible to use a symmetrization argument to reduce the problem to the symmetric case ?