Positiveness of energy of differential equation

This is a partial answer (but may put someone on the right track for a full one)

In the case $E$ and $h$ are $C^{\infty}$ (infinitely derivable, like it is often the case in physics) and if there is a constant $M$ that bounds ALL of the derivatives of $E$:

Let us assume $E(t_0) = 0$ for some $t_0$

Then, we can prove recursively that all the derivatives of $E$ are null in $t_0$ by differentiating $E'= h \times E $ (you always get some $E^{(k)}$ in factor of the terms, which means $E^{(n+1)}(t_0) = 0$)

Let us use Taylor's inequality in $t_0$ (https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder), we can have

$\forall n \in \mathbb{N}, E(0) \leq M\frac{t_0^n}{n!}$

Which, at its limit, gives

$E(0) \leq 0$, which is absurd.

Hence, $\forall t \in [0,T], E(t)>0$

Winther was right. You can use Gronwall's Lemma to understand it. Define the function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that: $$f(t)=E(x(t))^2\times 1_{E(x(t))\leqslant 0}$$ For sure, this is a non-negative function. Take the derivative: \begin{align} \partial_t f(t)&\leqslant h(x(t),t)E(x(t))^2 \times 1_{E(x(t))\leqslant 0} \\ &\leqslant h(x(t),t)f(t) \end{align} this implies that: $$f(t) \leqslant \int_0^t h(x(s),s) f(s) ds $$ By hypothesis $(E(x(0))>0 \Leftrightarrow f(0)=0)$, which gives: $$f(t) \leqslant f(0) \times \int_0^t h(x(s),s) f(s) ds=0 $$ By definition of $f$, this implies that $E(x(t))$ is positive. Of course, here all the $\leqslant$ could be replaced by $=$ since you have equalities. Note that the derivative of $1_{E(x(t),t)\leqslant 0}$ has a meaning with the notion of weak derivative; its "derivative" is a dirac at the point "$E(x(t),t)=0$".