For the square root function $g(x) = \sqrt{x}$, we have
\begin{align} g'(x) = 0.5x^{-0.5} \\ g''(x) = -0.25x^{-1.5} \\ \end{align}
So the second derivative is negative everywhere except if $x = 0$. So $g(x)$ is strictly concave if $x > 0$. But when $x = 0$, the second derivative is undefined. What can we say about the strict concavity or the square root function at $x = 0$?
It seems that we can't prove strict concavity using the second derivative, but using Jensen's inequality, I think we can fairly easily show it, though even with Jensen's, we would have to assume that $t \in (0, 1)$ instead of $t \in [0, 1]$, or we run into the same issue.