Sufficient condition for $ \sigma_{X}^{2} \leq \sigma_{Y}^{2}$ Suppose $X$ and $Y$ are random variables whose expected values are $\mu_X$ and $\mu_Y$, and variances are $\sigma_{X}^{2}$ and $\sigma_{Y}^{2}$, respectively.
Also, we suppose $F_x$ and $F_Y$ are the distribution function of $X$ and $Y$.
Say that the inquality
$$ F_{X}(\mu_X+t) - F_{X}(\mu_X-t) \geq F_{Y}(\mu_Y+t) -  F_{Y}(\mu_Y-t)$$
holds for all $t$.
Then, is it true to be
$$ \sigma_{X}^{2} \leq \sigma_{Y}^{2}$$?
The only thing that I was able to do is deriving the inequalities below, but I don't think these work for the problem.
\begin{align*}
\sigma_{X}^{2} 
&\geq t^2 \left\{ F_{X}(\mu_X - t) - F_{X}(\mu_X + t) \right\} \\
\sigma_{Y}^{2} 
&\geq t^2 \left\{F_{Y}(\mu_Y - t) - F_{Y}(\mu_Y + t)\right\}
\end{align*}
I would appreciate if you could help me.

Let us assume, without loss of generality, that $\mu_X = \mu_y = 0$.  We then have:
$$F_X(t)  - F_X(-t)  \geq F_Y(t) - F_Y(-t)$$
We observe that $\sigma^2_X = \mathbb{E}x^2$ (as we have set the expectation equal to zero) and similarly for $\sigma^2_Y$; therefore we want to work with squares instead of the initial variables.   We denote the cumulative distributions of $x^2$ and $y^2$ by $F^{'}_X$ and $F^{'}_Y$,and define $t^{'} = t^2$.  Since $x^2 \leq $ any arbitrary $c^2$ is equivalent to $-c \leq x \leq c$, it's clear that $F^{'}_X(t^{'}) = F_X(t) - F_X(-t)$, which leads directly to:
$$F^{'}_X(t^{'}) \geq F^{'}_Y(t^{'})$$
Now, as $t^{'}$ is non-negative, being a square, we know that $\mathbb{E}t^{'} = \int_0^{\infty}(1-F_{t^{'}}(t^{'}))dt^{'}$.  Substituting and engaging in some minor algebra gives us:
$$\sigma_Y^2 - \sigma_X^2 = \int_0^{\infty}(1-F_Y^{'}(y))dy - \int_0^{\infty}(1-F_X^{'}(x))dx$$
which can be rearranged as:
$$\sigma_Y^2 - \sigma_X^2 = \int_0^{\infty}(F^{'}_X(t) - F^{'}_Y(t)) dt$$
Since $F^{'}_X(t) - F^{'}_Y(t) \geq 0$ for all $t$, the integral on the right is bounded below by $\int_0^{\infty}0dt = 0$; therefore:
$$\sigma_Y^2 - \sigma_X^2 = \int_0^{\infty}(F^{'}_X(t) - F^{'}_Y(t)) dt \geq 0$$
and $\sigma^2_X \leq \sigma^2_Y$.
