If I prove the estimator of $\theta^2$ is unbiased, does that prove that the estimator of parameter $\theta$ is unbiased? Let $X_i$ be an iid random variable having pdf $f(\mathbf{x}|\theta)$, where $E(X_i) =  6\theta^2$, and $\theta > 0$.
I have calculated an estimator for the parameter ($\theta$) of $f(\mathbf{x}|\theta)$ to be $\hat{\theta} = \sqrt{\bar{x}/6}$. To prove that this is an unbiased estimator, I should prove that $E(\hat{\theta}) = E\left(\sqrt{\bar{x}/6}\right)$. However, since $\hat{\theta}^2 = \bar{x}/6$, it would be much easier to show that $$\begin{align}  E(\hat{\theta}^2) &= E(\bar{x}/6) \\ &=\frac{1}{6}E\left(\frac{\sum X_i}{n}\right)\\
&=\frac{1}{6n}\sum E(X_i) \\
&=\frac{1}{6n}n6\theta^2 \\&= \theta^2.\end{align}$$
Generally, proving $x^2 =4$ is not the same as proving $x=2$, since $x$ could also be $-2$. However, in this case $\theta>0$.
I have shown that $\hat{\theta}^2$ is unbiased, is this sufficient to show that $\hat{\theta}$ is unbiased?
 A: Note that for any estimator (with finite second moment) that $E(\widehat{\theta^2}) - E(\hat\theta)^2$ $=$ $\text{Var}(\hat\theta)\geq 0$ with equality only when $\text{Var}(\hat\theta)=0$ (which is easy to check doesn't hold). 
Replace the first term on the LHS of that inequality by using your result for unbiasedness of $\widehat{\theta^2}$, and then by using the fact that $\theta$ and $\hat \theta$ are both positive, show $\hat \theta$ is biased, not unbiased as you supposed. (More generally, you could apply Jensen's inequality but it's not needed here)
Note that this proof doesn't relate to the particulars of your problem -- for a non-negative estimator of a non-negative parameter, if its square is unbiased for the square of the parameter, then the estimator must itself be biased unless the variance of the estimator is $0$.
A: Say $Q$ is unbiased for $\theta^2$, i.e. $E(Q) = \theta^2$, then because of Jensen's inequality, 
$$\sqrt{ E(Q) } = \theta < E \left( \sqrt{Q} \right)$$
So $\sqrt{Q}$ is biased high, i.e. it will overestimate $\theta$ on average. 
Note: This is a strict inequality (i.e. $<$ not $\leq$) because $Q$ is not a degenerate random variable and square root is not an affine transformation. 
