Let $X_1, X_2,\cdots, X_n$ be normal random variables with mean $\mu=a$ and variance $\sigma^2=a$.
I want to show that $$E(\frac{\sqrt{1+4s}-1}{2})\neq a$$ where $s=\frac{\sum_{i=0}^nX_i^2}{n}$.
It seems exact calculation is difficult so maybe Jensen's inequality would be helpful.
Since $x\mapsto\sqrt x$ is a concave function, Jensen's inequality yields \begin{align} \mathbb E\left[\frac{\sqrt{1+4S_n}-1}2\right] &= \frac12(\mathbb E[\sqrt{1+4S_n}]-1)\\&\leqslant \frac12\left(\sqrt{\mathbb E[1+4S_n]}-1\right)\\ &=\frac12\left(\sqrt{1 + 4\mathbb E[S_n]}-1 \right)\\ &=\frac12\left(\sqrt{1+4a}-1\right). \end{align} Now, $$ a = \frac12\left(\sqrt{1+4a}-1\right) \iff a=0, $$ which is impossible since a normal random variable must have positive variance.