Finding tight upper/lower bounds for $\mathbb{E}[\frac{1}{1+X^{2}}]$ where $X$ is a RV with $\mathbb{E}[X]=0$ and $\mbox{Var}(X)=\nu<\infty $ The question is pretty much in the title.
My first thought was using Jensen's inquality to get some sort of lower bound. Since $\frac{1}{1+x^{2}}$
  is convex on $\mathbb{R}\backslash\left[-\frac{1}{2},\frac{1}{2}\right]$
  I think by Jensen's inequality I get a lower bound of the form: $$\frac{1}{1+\nu}\leq\frac{1}{1+\mathbb{E}\left[\left(X\cdot1_{\left\{ X>\frac{1}{2}\right\} }\right)^{2}\right]}\leq\mathbb{E}\left[\frac{1}{1+\left(X\cdot1_{\left\{ X>\frac{1}{2}\right\} }\right)^{2}}\right]\leq\mathbb{E}\left[\frac{1}{1+X^{2}}\right]
 $$
 I'm not sure if it's even correct to use Jensen's inequality in this manner, the formulation of the inequality I'm familiar with is for an integrable RV and a convex function over an interval of the form $\left(a,b\right)$
  with $a,b$
  possibly being $\pm\infty$. 
EDIT: like a comment suggested it's not hard to apply Jensen's Inequality correctly to obtain the second inequality. Unfortunately I just noticed the third inequality is incorrect since $$\frac{1}{1+\left(X\cdot1_{\left\{ X\geq\frac{1}{2}\right\} }\right)^{2}}\geq\frac{1}{1+X^{2}} $$
Yet the bound itself still feels like it should be correct. I also verified you can approach it as desired with a uniform RV defined on the interval $\left(-\varepsilon,\varepsilon\right)$ as $\varepsilon$ tends to $0$. As can be seen:
$$\mathbb{E}\left[\frac{1}{1+X^{2}}\right]=\int\limits _{-\varepsilon}^{\varepsilon}\frac{1}{2\varepsilon}\cdot\frac{1}{1+x^{2}}dx=\frac{\tan^{-1}\left(\varepsilon\right)}{\varepsilon}\overset{\varepsilon\to0}{\longrightarrow}1$$
$$\frac{1}{1+\mathbb{E}\left[X^{2}\right]}=\frac{1}{1+\frac{\varepsilon^{2}}{3}}\overset{\varepsilon\to0}{\longrightarrow}1$$
I'd appreciate help showing a bound of the form $\frac{1}{1+\nu}\leq\mathbb{E}\left[\frac{1}{1+X^{2}}\right]$ indeed holds (or a contradicting example and an alternative bound).

The function $t\mapsto\frac1{1+t}$ is convex on $t\geqslant0$ hence 
$$
E\left(\frac1{1+X^2}\right)\geqslant\frac1{1+E(X^2)}=\frac1{1+\nu}.
$$
The lower bound is attained when $P(X=\sqrt\nu)=P(X=-\sqrt\nu)=\frac12$.
On the other hand, if $P(X=0)=1-\frac\nu{x^2}$ and $P(X=x)=P(X=-x)=\frac\nu{2x^2}$ for some $|x|\geqslant\sqrt\nu$, then $E(X)=0$ and $E(X^2)=\nu$ while $$E\left(\frac1{1+X^2}\right)=1-\frac\nu{1+x^2},$$ hence the best universal upper bound valid for every $X$ such that $E(X)=0$ and $E(X^2)=\nu$ with $\nu\gt0$ is
$$
E\left(\frac1{1+X^2}\right)\lt1.
$$
