Variance of $(X-\mu)^2+\mu^2 $ smaller than the variance of $X^2$, but how much

Question

I have managed to prove that $Var(X^2) \ge Var((X-\mu)^2+\mu^2) $.

This is what I got with help from @Calvin Khor. $Var(X^2) - Var((X-\mu)^2+\mu^2) = 4E(X)E[X(X-E(X))^2]$

Now I wonder, with what factor is the variance of $(X-\mu)^2+\mu^2$ smaller than the variance of $X^2$?

Thanks in advance!

Answer 1

Hint : Variance or $\sigma^2$ is $E\left[(X-\mu)^2\right]$ and do a little algebra we get $E(X^2)-(E(X))^2$, considering $\mu=E(X)$