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Let $c>1/2$ be an arbitrary big fixed constant. Can one prove that for all $t\geq 1$: $$\text{erf}\left(\frac{c}{t}\right) \ge \delta \, \min\left(1,\frac{c}{t}\right)$$

for some small constant $\delta>0$?

The error function $\text{erf}(z)$ is defined as

$$\text{erf}(z) = \frac2{\sqrt{\pi}}\int_0^z e^{-t^2}\mathrm dt.$$

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up vote 2 down vote accepted

Since the function $a:t\mapsto\mathrm{e}^{-t^2}$ is nonincreasing on $t\ge0$, $a(t)\ge a(1)$ for every $t$ in $(0,1)$. Integrating this inequality yields $\mathrm{erf}(z)\ge\delta z$ for every $z$ in $(0,1)$, with $\delta=2a(1)/\sqrt{\pi}=2\mathrm{e}^{-1}/\sqrt{\pi}$. On the other hand, if $z\ge1$, then $\mathrm{erf}(z)\ge\mathrm{erf}(1)\ge\delta$, hence you are done for every positive $c/t$.

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