Not quite right

First, $$\begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . ..\end{align}$$ Subtract: $$ \mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt $$ So it suffices to show that this integrand is positive. It is positive for $t>1$ (proof needed), so we establish $\mathrm{erf}(x) > \mathrm{tanh}(x)$ for $x > 1$.

First, $$\begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . \end{align}$$ Subtract: $$ \mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt $$ So it suffices to show that this integrand is positive. Now $(t-1)^2 \ge 0$, so $-t^2 \le 1-2t$ so $e^{-t^2} \le e \cdot e^{-2t}$. Also $\cosh t = (e^t+e^{-t})/2 < e^t/2$ so $\mathrm{sech}^2 t \gt 4e^{-2t}$. Once we note that $2 e/\sqrt{\pi} \approx 3.07 < 4$, we have $$ \frac{2}{\sqrt{\pi}} e^{-t^2} < \frac{2 e}{\sqrt{\pi}} e^{-2t} < 4 e^{-2t} \le \mathrm{sech}^2 t , $$ as required

Not quite right .

Conclusion: $\mathrm{erf} x < \mathrm{tanh} x$ for all real $x>0$...

First, $$\begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . \end{align}$$ Subtract: $$ \mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt $$ So it suffices to show that this integrand is positivefor $t > 0$. Now $(t-1)^2 \ge 0$, so $-t^2 \le 1-2t$ so $e^{-t^2} \le e \cdot e^{-2t}$. Also $\cosh t = (e^t+e^{-t})/2 < e^t/2$ so $\mathrm{sech}^2 t \gt 4e^{-2t}$. Once we note that $2 e/\sqrt{\pi} \approx 3.07 < 4$, we have $$ \frac{2}{\sqrt{\pi}} e^{-t^2} < \frac{2 e}{\sqrt{\pi}} e^{-2t} < 4 e^{-2t} \le \mathrm{sech}^2 t , $$ as required.

Conclusion: $\mathrm{erf} x < \mathrm{tanh} x$ for all real $x>0$.

First, $$\begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . \end{align}$$ Subtract: $$ \mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt $$ So it suffices to show that this integrand is positive for $t > 0$. Now $(t-1)^2 \ge 0$, so $-t^2 \le 1-2t$ so $e^{-t^2} \le e \cdot e^{-2t}$. Also $\cosh t = (e^t+e^{-t})/2 < e^t/2$ so $\mathrm{sech}^2 t \gt 4e^{-2t}$. Once we note that $2 e/\sqrt{\pi} \approx 3.07 < 4$, we have $$ \frac{2}{\sqrt{\pi}} e^{-t^2} < \frac{2 e}{\sqrt{\pi}} e^{-2t} < 4 e^{-2t} \le \mathrm{sech}^2 t , $$ as required.