This is the answer I worked out in
arXiv:1406.5909
Theorem
Let $y: [0,c] \to \mathbb{R}$, be any $C^n$ function which satisfies the differential inequality
$$
y^{(n)}+a_1(x) y^{(n-1)}+\cdots+ a_{n}(x) y\ge 0, \qquad(1)
$$
and the initial conditions
$$
y^{(i)}=0, \textrm{ for } i=0,1,\cdots,n-1,
$$
where the coefficients $\{a_i(x)\}$ are Lipschitz.
Then there is some constant $b$, $0<b\le c$, dependent on the coefficients $\{a_i(x)\}$ but independent of $y$, such that for every $y$ and every $i=0,\cdots, n-1$ we have $y^{(i)}\ge 0$ on $[0,b]$. Moreover, the equality $y=0$ holds on the whole interval $[0,b]$ if and only if the equality in (1) holds on the whole interval $[0,b]$.
Proof
Let us denote with $L$ the linear differential operator on the left-hand side of (1). Let $K(x,\xi)$ be the Cauchy function, that is the function defined on the triangle $0\le \xi \le x < c/2$ which solves $L K=0$ for every $\xi \in[0,c/2)$ with initial condition
$$
K(\xi,\xi)=K^{(1)}(\xi,\xi)=\cdots=K^{(n-2)}(\xi,\xi)=0, \quad K^{(n-1)}(\xi,\xi)=1,
$$
where these derivatives are with respect to $x$. This function is well defined because it is equivalently determined by the linear system of first-order differential equations in the dependent variables $(K^\xi_0, K^\xi_1, K^\xi_2,\cdots, K^\xi_{n-1}) : [0,c/2]\to \mathbb{R}$
\begin{align*}
\frac{d}{ds}{K^\xi}_0&=K^\xi_1, \\
\cdots &\\
\frac{d}{ds}{K^\xi}_{n-2}&= K^\xi_{n-1},\\
\frac{d}{ds} {K^\xi}_{n-1}&=- a_1(s+\xi) K^\xi_{n-1}-a_2(s+\xi) K^\xi_{n-2}\cdots-a_n(s+\xi) K^\xi_0 .
\end{align*}
where $s+\xi=x$, under the initial condition
$$
K^\xi_0(0)=K^\xi_1(0)=\cdots=K^\xi_{n-2}(0)=0, \quad K^\xi_{n-1}(0)=1.
$$
Indeed, there is one and only one solution by the Picard-Lindel\"of theorem (Cor. 5.1 in Hartman 1964 Ordinary Differential Equations), and so we obtain our desired Cauchy function once we set $K(x,\xi)=K^\xi_0(x-\xi)$ (thus $K^{(i)}(x,\xi)=K^\xi_i(x-\xi)$). Moreover, the above system of first-order differential equations is Lipschitz also with respect to the external parameter $\xi$, thus its solution $(K^\xi_0, K^\xi_1, K^\xi_2,\cdots, K^\xi_{n-1}) $ has a Lipschitz dependence on $(s,\xi)$ (this is Peano's theorem; if the $a$s are $C^1$ then one can infer that the $K$s are $C^1$ too (Theor. 3.1 in Hartman (1964) Ordinary Differential Equations), for the Lipschitz case see (Lang 1995, Differential and Riemannian manifolds or Cartan 1971 Differential Calculus). The function $K^{(n-1)}(x,\xi)$ is continuous in both $(x,\xi)$ on the triangle $0\le \xi \le x < c/2$ and in particular at $(0,0)$. Since $K^{(n-1)}(0,0)=1>0$, there is some neighborhood of $(0,0)$ (in the product topology) over which $K^{(n-1)}$ is positive, and hence a triangle $0\le \xi \le x < b\le c/2$ over which $K^{(n-1)}$ is positive. But on the diagonal $K^{(i)}$, $i=1,\cdots, n-2$, vanishes so upon integration on $x$ we obtain that $K^{(i)}$, is positive on $0\le \xi < x < b$ for every $i=1,\cdots, n-1$.
From the assumption we have that $Ly\ge 0$, where $Ly$ is continuous. The uniqueness of the solution to the differential equation $L y= f$ implies the easily verifiable formula
$$
y(x)=\int_0^x K(x,\xi) Ly(\xi) \, d \xi ,
$$
which under differentiation gives more generally
$$
y^{(i)}(x)=\int_0^x K^{(i)}(x,\xi) Ly(\xi)\, d \xi , \qquad i=0,1,\cdots, n-1,
$$
thus $y^{(i)}\ge 0$ on $[0,b]$, and the equality $y=0$ on $[0,b]$ is possible only if $Ly=0$ on $[0,b]$. $\square$
