How are $Var(X)$ and $Var(X^2)$ both positive? 
Suppose that the distribution of a random variable
  $X$ is symmetric with respect to the point $x = 0$.  If $\mathbb{E}(X^4)>0$ then $Var(X)$ and $Var(X^2)$ are both positive.

How is that true? I am getting $Var(X)=\mathbb{E}(X^2)$ and  $Var(X^2)=\mathbb{E}(X^4)-(\mathbb{E}(X^2))^2$, but do not know why $\mathbb{E}(X^2)>0$ & $\mathbb{E}(X^4)>(\mathbb{E}(X^2))^2.$
 A: I do not think it is true.  
For example let $X=k$ and $X=-k$ each have probability $\frac12$ for some $k\gt 0$.
Then the distribution is symmetric about $0$, i.e. $P(X \le -x) =P(X \ge x)$ for all $x$.
And $E[X]=0$, $E[X^2]=k^2 \gt 0$ and $E[X^4]=k^4 \gt 0$, and $Var(X)=k^2 \gt 0$.
But  $Var(X^2)=0$, contrary to the statement in the question.
A: We know that the variance of a random variable is $0$ if and only if it is constant (so $X = \text{E}(X)$).

Proof:
  If $X$ has constant value $c$, then $\text{E}(X) = c$ and $\text{E}(X^2) = \text{E}(c^2) = c^2$, $\Rightarrow \text{V}(X) = \text{E}(X^2) - \text{E}(X)^2 = 0$.
  Conversely, suppose the variance, $\sigma^2 = 0$, and let the mean be $\mu$. Then by Chebyshev's inequality
  $\Pr[|X - \mu| \ge k] \le \dfrac{\sigma^2}{k^2} \le 0$, so that $|X - \mu| = 0 \Rightarrow X = \mu$.

As $X$ is symmetric about $0$, the mean is $\text{E}(X) = 0$. But as $\text{E}(X^4) > 0$, $X^4$ has some non-zero values, which implies that $X$ and $\not X^2$ also have has some non-zero values. Hence, $X$ and $\not X^2$ are is not constant. Thus, $\text{V}(X) > 0$ and $\not{\text{V}(\not X^2)} \not > \not 0$.
Edit: Henry is right, this does not imply that $V(X^2) > 0$.
A: I think we can prove everything but the fact that $Var(X^2) > 0.$ We are given that $E(X) = 0$ and $E(X^4) > 0$, the question is how to show both Var$(X)$ and Var$(X^2)$ are strictly positive.
Knowing $E(X^4) > 0$ allows us to say that
$$
\begin{align}
\frac{\sum_{i=1}^nx_i^4}{n} &> 0\\
\Rightarrow \sum_{i=1}^nx_i^4 &> 0
\end{align}
$$
Now each $x_i^4$ individually must be greater than or equal to $0$ as the power is even, and there must exist some $x_i \neq 0$ to allow for $\sum_{i=1}^nx_i^4 > 0$. Discard all $x_i = 0$, and call the remaining observations the set ${i^*}$. The remaining $x_{i^*} \neq 0$ so $\sum_{i^*} x_{i^*}^2 > 0$ as well. All that are left are 0 observations which cannot reduce the value, so $E(X^2)$ must also be greater than 0, thus:
$$
\begin{align}
Var(X) &= E(X^2) - E(X)\\
&= E(X^2) > 0\\
&\blacksquare\; Var(X) > 0
\end{align}
$$
Now we get to the tricky part. We can say
$$
\begin{align}
Var(X^2) &= E((X^2)^2) - E(X^2)^2\\
&= E(X^4) - E(X^2)^2
\end{align}
$$. 
What is $E(X^4)$? It's
$$
\begin{align}
&=\frac{\sum_{i^* = 1}^{n^*}x_{i^*}^4}{n^*}\\
&=\frac{\sum_{i^* = 1}^{n^*}\left(x_{i^*}^2\right)^2}{n^*}\\
\end{align}
$$
What is $E(X^2)^2$? It's
$$
\begin{align}
&=\left(\frac{\sum_{i^* = 1}^{n^*}x_{i^*}^2}{n^*}\right)^2\\
&=\frac{\left(\sum_{i^* = 1}^{n^*}x_{i^*}^2\right)^2}{\left(n^*\right)^2}
\end{align}
$$
So the denominator of $E(X^2)^2$ is the square of the denominator of $E(X^4)$ which should make $E(X^4) > E(X^2)^2$. However, compare the numerators. We have a case of Jensen's inequality. Since all the terms are strictly non-negative, and some are strictly positive, being that squaring is a convex function, the sum of the squares is less than or equal to the square of the sum. So we have $E(X^4)$ having a lower numerator and $E(X^2)^2$ having a higher denominator, and I'm not sure if we can always say which one dominates.
