Show that $\int_Ax^2\mu_X(dx)\le\frac{12}{11u^2}\{1-\Re(\Phi_X(u))\}$ 
Let $A=[-\frac1u,\frac1u]$, Show that $$\displaystyle\int_Ax^2\mu_X(dx)\le\frac{12}{11u^2}\{1-\Re(\Phi_X(u))\}$$
where $\Phi_X(u)$ is the characteristic function of the r.v. $X$
Hint: $1-\cos(x)\ge0$  and $1-\cos(x)\ge\frac{x^2}{2}-\frac{x^4}{24}$

$\displaystyle\{1-\Re(\Phi_X(u))\}=1-\Re\bigg(\int e^{iux}\mu_X(dx)\bigg)=1-\int\cos\langle u,x\rangle\mu_X(dx)$
$\displaystyle=\int 1-\cos\langle u,x\rangle\mu_X(dx)\ge\int_{|x|\le\frac{\sqrt{12}}{u}}\frac{u^2x^2}{2}-\frac{u^4x^4}{24}\mu_X(dx)$
(because  $1-\cos(x)\ge0$ and therefore we need only the positive part of $\frac{x^2}{2}-\frac{x^4}{24}$, which is in $\{|x|\le\frac{\sqrt{12}}{u}\}$)
Now I have to estimate correctly, but I don't see how, I tried to separate the integral, for the first I took the domain $\{|x|\le\frac{1}{u}\}$, which gave an integral greater than $\int_{\{|x|\le\frac{1}{u}\}}\frac{x^2}{2}$, but I couldn't manage to do an estimation for the second one, for the domain $\{\frac{1}{u}\le|x|\le\frac{\sqrt{12}}{u}\}$)
Change of variable also failed, I even don't know if it's allowed to replace $x$  without changing $\mu_X(dx)$
Any suggestions would be appreciated.
Thanks.
 A: All the pieces are there, this is only a matter of using them in the right order. Note first that
$$
1-\Re\varphi_X(u)=\Re(1-\varphi_X(u))=\int_\mathbb R(1-\cos(ux))\mu_X(\mathrm dx),
$$
hence
$$
1-\Re\varphi_X(u)\geqslant\int_A(1-\cos(ux))\mu_X(\mathrm dx),
$$
because $1-\cos$ is nonnegative everywhere. Next, note that, for every $|t|\leqslant1$,
$$
1-\cos(t)\geqslant\frac12t^2-\frac1{24}t^4\geqslant\frac{11}{24}t^2,
$$
because $t^4\leqslant t^2$ when $|t|\leqslant1$, hence the second inequality provided for $1-\cos$ yields, for every $x$ in $A$,
$$
1-\cos(ux)\geqslant\frac{11}{24}u^2x^2,
$$
which implies
$$
\int_A(1-\cos(ux))\mu_X(\mathrm dx)\geqslant\frac{11}{24}u^2\int_Ax^2\mu_X(\mathrm dx).
$$
This proves that

$$
\int_Ax^2\mu_X(\mathrm dx)\leqslant\frac{24}{11u^2}(1-\Re\varphi_X(u)).
$$

Note that a limited expansion of $\varphi_X(u)$ when $u\to0$ shows that every factor smaller than $2$ (such as your $\frac{12}{11}$) is impossible if $\varphi_X$ is defined as
$$
\varphi_X(u)=E[\mathrm e^{\mathrm iuX}].
$$
Assume for example that $X=\pm1$ with equal probability, then, for every $|u|\leqslant1$,
$$
\int_Ax^2\mu_X(\mathrm dx)=1,
$$
while
$$
\varphi_X(u)=\cos u,
$$
hence, when $u\to0$,
$$
\frac1{u^2}(1-\Re\varphi_X(u))=\frac1{u^2}(1-\cos u)\to\frac12.
$$
