Jensen inequality implies that for every real random variable $X$ and every integer $n\in \mathbb N$ $$ (\mathbb E[X^2])^n \,\leq\, \mathbb E[X^{2n}]$$ by convexity of the function $x\mapsto x^n$ for $x\geq0$.
Now if we apply the previous inequality to a Gaussian random variable $X$ of mean $0$ and variance $\sigma^2$ we find: $$ \sigma^{2n} \,\leq\, \sigma^{2n}\, (2n-1)!!$$ which of course is not very good for large $n$ since $(n-1)!!$ grows faster than exponential.
Is there an alternative / improvement of Jensen inequality that bounds $(\mathbb E)^n$ with a linear operator defined on a suitable space of random variables functions of $X^2$ and preserving exponential growth for large $n$?
Edit. I would be happy enough if I could do that for all $X$ having a strong log-concave distribution on $\mathbb R$, namely $$\mathbb E [\varphi(X)] \,=\, \int_{-\infty}^{\infty}\, \varphi(x)\,e^{-U(x)} \,dx $$ with $U''(x)\geq\lambda>0$ for all $x\in\mathbb R\,$, for all $\varphi:\mathbb R\to\mathbb R$ having polynomial growth. Notice that the Gaussian case corresponds to $U(x)=(x-\mu)^2/(2\sigma^2) + \frac{1}{2}\log(2\pi\sigma^2)$.
Edit 2. A possible modification of this question could be: for which family of random variables it holds true $$ (\mathbb E[X^2])^n \,\leq\, \frac{c^n}{(2n-1)!!}\,\mathbb E[X^{2n}]$$ for some $c>0$ and all $n\in\mathbb N\,$? Centred Gaussian random variables for example satisfy the inequality with $c=1$.