# Proving that a function in an ODE has an asymptote

So I've looked at this answer to the problem of showing that the function $$y$$ that satisfies: $$y'=1+y^4$$ has an asymptote. The solution seems very elegant, except I cannot follow one of the steps. Namely, when he goes from $$\frac{y'}{y^2}\geq2$$ to: $$\frac{1}{y(t_0)}-\frac{1}{y(t_1)}\geq2(t_1-t_0)$$ I understand that since $$y$$ is (obviously) concave up, the slope of the secant from $$t_0$$ to $$t_1$$ is larger than the derivative at $$t_0$$: $$y'(t_0)\leq \Delta y=\frac{y(t_1)-y(t_0)}{t_1-t_0}$$ Which means that we can bound $$\frac{y'}{y^2}$$ with: $$\frac{\Delta y}{y^2}\geq\frac{y'}{y^2}\geq2$$ And since $$y^2=y(t_0)^2$$, we may write: $$\frac{y(t_1)-y(t_0)}{y(t_0)^2}\geq2(t_1-t_0)$$ However, this is not the same inequality that is derived in the answer. I realize that the answer's inequality may be derived if we let $$y^2\approx y(t_0)y(t_1)$$, and this approximation becomes more accurate if we bring $$t_1$$ closer and closer to $$t_0$$.

So my question is, how is that particular inequality derived?

Integrate $$\frac{y'}{y^2}\geq 2$$ from time $$t=t_0$$ to $$t=t_1$$, $$t_0 gives $$\int_{t_0}^{t_1}\frac{y'}{y^2}\,\mathrm{d}t \geq \int_{t_0}^{t_1}2\,\mathrm{d}t$$ and $$LHS=\int_{y(t_0)}^{y(t_1)}\frac{\mathrm{d}y}{y^2}=\frac1{y(t_0)}-\frac1{y(t_1)}.$$