<s>[INST] Why is $\operatorname{Var}(X^2) \ge\operatorname{Var}(X − \mu)^2 + \mu^2)$? I wonder why $\operatorname{Var}(X^2) \ge \operatorname{Var}((X − \mu)^2 + \mu^2)$.
IN ADDITION
What factor is the variance of $(X − \mu)^2 + \mu^2$ smaller than the variance of $X^2$?
What I've done so far:
$$\operatorname{Var}((X − \mu)^2 + \mu^2) =\operatorname{Var}(X^2 -2X\mu+2\mu^2) = \operatorname{Var}(X^2 -2X\mu).$$
From here, I wonder if you have any ideas?
 [/INST] The result is only conditionally true (I assume you mean $\mu = E(X)$). It is valid assuming that either $E(X) = 0$, $X\ge 0$ or $X\le 0$. If no condition is given, it does not always hold, and we prove this by a simple counterexample below.
$\operatorname{Var}X = E(X^2) - E(X)^2$. Sub in
\begin{align} \mathbf \Delta(X) &:=\operatorname{Var}(X^2) - \operatorname{Var}((X-\mu)^2 + \mu^2)\\ &\phantom{:}= E(X^4) - E(X^2)^2 - E((X^2-2X\mu )^2) + (E(X^2-2X\mu))^2.\end{align}
We have by direct computation,
\begin{align} E((X^2-2X\mu )^2) &= E(X^4) - 4\mu E(X^3) + 4\mu^2E(X^2),\\
  (E(X^2-2X\mu))^2 &= E(X^2)^2 - 4\mu E(X)E(X^2) + 4\mu^2E(X)^2.\end{align}
thus
$$ \mathbf \Delta(X)  = 4\mu E(X^3) - 4\mu^2 E(X^2) - 4 \mu E(X) E(X^2) + 4\mu^2 E(X)^2.$$
Setting $\mu = E(X)$,
\begin{align} \mathbf \Delta(X)   &= 4 \mu E(X^3) - 8 \mu^2 E(X^2) + 4\mu^4 \\ 
&= 4\mu [ E(X^3) - 2 \mu E(X^2) + \mu^2E(X)] \\
&=4\mu  E(X^3 - 2 \mu X^2 + \mu^2 X) \\
& = 4\mu E\Big (X ( X-\mu)^2 \Big) = 4E(X) E\Big (X ( X-E(X))^2 \Big)
\end{align}
Suppose that $X \ge 0$. Then $E(X)\ge 0$. Also, $X ( X-E(X))^2\ge 0$, so $E(X ( X-E(X))^2) \ge 0$, and therefore the inequality $\mathbf\Delta(X) \ge 0$ holds for $X\ge 0$. But if $X\le 0$, then setting $Y=-X\ge 0$, since $X ( X-E(X))^2 = -Y(Y-E(Y))^2$, we also have
$$  \mathbf \Delta (X) = \mathbf \Delta (Y) \ge 0.$$

Now we give a simple counterexample that shows it cannot hold in general. Let $X$ be the following random variable that only takes the values $\pm 1$,
$$ P(X=1) = 1-P(X=-1) = p_1 \notin \{0,1/2,1\}.$$
Then as $X^2 \equiv 1$ is almost surely constant, $\operatorname{Var}(X^2) = 0$. But the right-hand side is randomly varying since $p_1 \neq 0,1$, and we avoided the sneaky cancellation from $E(X)$ by choosing $p_1 \neq 1/2$. Thus $\operatorname{Var}((X-\mu)^2 + \mu^2)>0$, and therefore
$$ \mathbf \Delta(X) < 0.$$
Concretely, with $p_1 \approx 0.853$, we have $\mathbf \Delta(X) \approx -0.99999021$.
</s>