I'm trying to prove the following inequality for $0 < x < 1$:

$$\operatorname{erf}\left(\frac{(1+x)\sqrt{\ln{(1+x)}}}{\sqrt{(1+x)^2 - 1}}\right) - \operatorname{erf}\left(\frac{\sqrt{\ln{(1+x)}}}{\sqrt{(1+x)^2 - 1}}\right) \geq \frac{x}{4}$$

Proof by WolframAlpha: http://goo.gl/15mrM

I could also construct a Proof by Mathematica, without too much trouble.

However, I'm looking for a more elegant proof of this inequality. My approach was going to involve showing that this holds for $x = 0$ and $x = 1$, and then show the function is concave. However, taking the second derivative yields the following monstrosity: http://goo.gl/fKxca

Is there a more elegant way to prove this? I wouldn't mind showing a weaker inequality of the form $\geq \frac{x}{c}$ (for some explicit $c$) if the proof was sufficiently simple.

  • $\begingroup$ I agree that erf(x) is concave. However, the difference of two concave functions is not necessarily concave. Consider the first function as y = -0.5x^2, and the other as y = -x^2. The difference will be y = 0.5x^2, which is convex. $\endgroup$ Jul 17, 2013 at 2:57
  • $\begingroup$ @newzad: as Gautam pointed out, it isn't because $f, g$ are concave that the difference of the two is (in other terms for $\mathcal ¢^2$ functions, $f'', g'' \leq 0 \not\Rightarrow f''-g'' \leq 0$, clearly). Similarly, the problem is not erf, but the error function composed with another function; and again, in general, the concavity of $f\circ g$ does not follow from the concavity of $f$ (or even of $f$ and $g$). $\endgroup$
    – Clement C.
    Jul 17, 2013 at 9:26

1 Answer 1


I don't know if it's the elegant approach you're looking for, but here's a suggestion: fix any $x_0 > $, and define $$ f_{x_0}\colon y > 0 \mapsto \operatorname{erf}\left(\frac{(1+y)\sqrt{\ln(1+x_0)}}{\sqrt{(1+x_0)^2-1}}\right) $$ Now, what you want to prove is $$\begin{equation} f_{x_0}\!(x_0)-f_{x_0}\!(0) \geq \frac{x_0}{4} \tag{1} \end{equation}$$ so it is sufficient to prove that for all $y\in[0,2x_0]$, $$\begin{equation} f_{x_0}\!(y)-f_{x_0}\!(0) \geq \frac{y}{4} \end{equation}$$ i.e. $$\begin{equation} \frac{f_{x_0}\!(y)-f_{x_0}\!(0)}{y} \geq \frac{1}{4} \tag{2} \end{equation}$$ Since $f_{x_0}$ is concave, you can use the usual arguments about concavity/convexity (eg, a concave function has a decreasing slope).

Does that make sense? (I'm not sure it is easy, but the whole point is "just" to reduce the problem to an actual concave function — for which (2) might be easier))


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.