Let $x$ be a differentiable function on $\mathbb{R}$. I want to prove that for any time $t \geq t_0$ \begin{equation} \frac{1}{2} D^{\alpha} x^2(t) \leq x(t) D^{\alpha} x(t), \ \ \forall \alpha \in (0, 1), \end{equation} where $D^\alpha$ is the Caputo derivative. This is equivalent to showing that

$ x(t) D^{\alpha} x(t) - \frac{1}{2} D^{\alpha} x^2(t) \geq 0, \ \ \forall \alpha \in (0, 1)$.

After simplifying we have: \begin{equation}\label{11.67} \biggl[\frac{[x(t) - x(t_0)]^2}{2\Gamma (1- \alpha) (t- t_0)^\alpha}\biggr]+ \frac{\alpha}{2\Gamma (1- \alpha)} \int_{t_0}^{t} \frac{[x(t)-x(\tau)]^2}{(t-\tau)^{\alpha+1}} \,d\tau \geq 0. \end{equation}

If we prove that the function $f(\tau)=\frac{[x(t)-x(\tau)]^2}{(t-\tau)^{\alpha+1}}$ is integrable, then the theorem is proved. Can this result be obtained?