The sum of two symmetric random variables is symmetric If $X$ and $Y$ are two random variables with probability density functions which are symmetric around their respective means, their sum, $X+Y$, has a probability density function which is symmetric around its mean as well.
Could someone offer a proof outline? Thanks.

Edit:
whuber's example (see comments) shows that the simply specifying symmetric marginals does not result in a symmetric sum
Dilip Sarwate's response gives two conditions, each of which is sufficient by itself: circular symmetry of the joint and independence of the marginals
 A: This looks a lot like a homework exercise but nonetheless, here goes.
If $X$ and $Y$ are zero-mean independent random variables, then (assuming that
they are continuous random variables) we have that for any $z$,
$f_{X+Y}(z)$ is given by the convolution of the marginal densities. Thus,
$$\begin{align}
f_{X+Y}(z) &= \int_{-\infty}^\infty f_X(z-y)f_Y(y)\,\mathrm dy \tag{1}\\
&= \int_{-\infty}^\infty f_X(y-z)f_Y(-y)\,\mathrm dy, 
&\text{symmetry of the densities}\\
&= \int_{-\infty}^\infty f_X(-t-z)f_Y(t)\,\mathrm dt, 
&\text{substitution: } t = -y\\
&= \int_{-\infty}^\infty f_X((-z)-t)f_Y(t)\,\mathrm dt,\\
&= \int_{-\infty}^\infty f_X((-z)-y)f_Y(y)\,\mathrm dy,
&\text{substitution: } t = y \tag{2}\\
&= f_{X+Y}(-z) &\text{on comparing (1) and (2)}\tag{3}
\end{align}$$
If $X$ and $Y$ have nonzero means $\mu_X$ and $\mu_Y$ respectively
and their densities are symmetric about their respective
means, then $\hat{X} = X-\mu_X$ and $\hat{Y} = Y - \mu_Y$ can be used
in the above proof to show that $\hat{Z} = \hat{X} + \hat{Y} =
(X+Y) - (\mu_X+\mu_Y) = Z - \mu_Z$ has a density symmetric about
$0$, and so $Z$ has a density symmetric about $\mu_Z$. Or,
we can use the outline suggested in @Quantibex's answer to incorporate
the means in the proof itself.
Similar proofs can be written for discrete random variables.
While the result is always true for independent random variables,
it can  hold for some dependent random variables as well. As
an example, see this recently-closed question where it is shown that if $(X,Y)$ are
uniformly distributed on the unit disc (and hence have symmetric marginal
densities but are not independent), then $X+Y$ also has a symmetric 
density; in fact,
$$f_{X+Y}(z) = \frac{1}{\sqrt{2}}f_X\left(\frac{z}{\sqrt{2}}\right). \tag{4}$$
Indeed, $(4)$ is true whenever $(X,Y)$ have a
circularly symmetric joint density (uniform density, as in the closed
question, is not needed).  Another nice example (with nonzero means)
is the joint density that has value $2$ on the trapezoidal
region with vertices $(0,0), (1,1), (\frac 12, 1), (0,\frac 12)$ and
on the triangular region with vertices $(\frac 12,0), (1,0), (1,\frac 12)$.
It is readily verified that $X$ and $Y$ have
marginal densities $U(0,1)$ that are symmetric about their
mean $\frac 12$, and that they are not independent.
Nonetheless, the density of their sum is the convolution of the
marginal densities and is symmetric about $1$.
A: An outline of a proof (in the case where $X$ and $Y$ are independent) is the following.
Denote $f_X$ and $f_Y$ the density functions of $X$ and $Y$, and $\mu_X$ and $\mu_Y$ their respective means.
Note that $f_X(\mu_X + x) = f_X(\mu_X - x)$ for all $x$ by the symmetry of $f_X$, and similarly $f_Y(\mu_Y + y) = f_Y(\mu_Y - y)$ for all $y$.
Let $Z = X + Y$, and denote $f_Z$ its density function and $\mu_Z$ its mean. Obviously, $\mu_Z = \mu_X + \mu_Y$.
It can be shown that $f_Z$ is the convolution $f_X * f_Y$ of $f_X$ and $f_Y$, where
$$
(f_X * f_Y)(z) = \int_{-\infty}^\infty f_X(z - y)f_Y(y) dz .
$$
To prove the symmetry of $f_Z$, show that $f_Z(\mu_Z + z) = f_Z(\mu_Z - z)$ for all $z$ using the convolution $f_X * f_Y$, with an appropriate change of variable such as $y = t + \mu_Y$, and using the symmetry properties of $f_X$ and $f_Y$.
