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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?

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

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