Strict (or lack therefore) concavity of square root function 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.
 A: Another view:
One definition of concavity is as follows:
Suppose $a < b$. The function $g$ is [strictly] concave on the interval $[a,b]$ if for all $x$ with $a < x < b$ we have:
$$\frac {g(x) - g(a)}{x - a} > \frac{g(b) - g(a)}{b - a}.$$
For the function $g$ defined by $g(x) = \sqrt x$ and the interval $[0,b]$ we have for $0 < x < b$,
\begin{align}
\frac {g(x) - g(0)}{x - 0} &= \frac {\sqrt x}{x}, \\
&= \frac{1}{\sqrt x}, 
\end{align}
and similarly
\begin{align}
\frac {g(b) - g(0)}{b - 0} &= \frac {\sqrt b}{b}, \\
&= \frac{1}{\sqrt b}. 
\end{align}
Comparing, we see that for all $0 < x < b$,
$$\frac{1}{\sqrt x} > \frac{1}{\sqrt b},$$
or
$$\frac {g(x) - g(0)}{x - 0} > \frac {g(b) - g(0)}{b - 0}.$$
Thus, $g$ is strictly concave on $[0,b]$. As this is true for any $b > 0$, $g$ is strictly concave on $[0,\infty)$.
A: hint
Let $ a>0$.
$f:x\mapsto \sqrt{x} $ is strictly  concave at $ [0,a]$ because it is syrictly concave at $(0,a] $ and
$$(\forall y\in (0,a])\; (\forall t\in(0,1))\;\;$$
$$ f(ty)=\sqrt{ty}>t\sqrt{y}$$
since $\sqrt{t}>t$.
