Can anyone prove that $$\mathrm{avg}\left(\frac {a_i}{\left(1+a_i\right)^2}\right) \ge \frac{\mathrm {avg}({a_i})}{\left(1+\mathrm{avg}{(a_i)}\right)^2}$$ for a sequence of positive valued elements $a_i, \quad\text{with}\quad i= 1,2,\ldots,N$?

  • $\begingroup$ Unless I went wrong, the inequality does not seem valid everywhere. Don't you have more constraint on the $a_i$? $\endgroup$ – Laurent Duval Nov 1 '20 at 21:05

You can recognize the function $$f: x\mapsto \frac{x}{(1+x)^2}\,.$$

The question can be rephrased as: how do compare the average of the function values $\overline{f(a_i)}$ with the function of the average value $f(\overline{a_i})$? Graphically, this can be illustrated by the following graph (spoiler: from Behold! Jensen's inequality). Here, the average is represented by $\mathbb{E}$, the statistical expectation, akin to the classical average.

Jensen inequality on a parabola

The intuition here is the following: for a convex function, as the above, the average of the $y=f(x)$ ($\mathbb{E}[f(x)]$) is higher than the function of the averages of $x$ (or $f(\mathbb{E}[x])$).

In the discrete form, Jensen's inequality can be proven by induction Proof 1 (finite form). This inequality is reversed for a concave function. The reverse inequality is the one you are interested in. Now, function $f(x)$ is not concave on $[0,\infty[$, graph from RechnerOnline:

function rechner online

With $a_1=0$ and $a_1=1$, we get $$\overline{f(a_i)}-f(\overline{a_i}) = 1/8 - 2/9 \le 0 \,.$$

With $a_1=1$ and $a_1=9$, we get $$\overline{f(a_i)}-f(\overline{a_i}) = 17/100 - 5/36 \ge 0 \,.$$

Hence, the inequality has a sign change. It would be valide for instance on some interval $[\alpha,+\infty)$, such that $f$ is concave.

  • $\begingroup$ Thank you very much for your help. You are right. The assumption is ai >> 1 to make this lower bound valid. $\endgroup$ – Xiaojing Huang Nov 2 '20 at 2:58
  • $\begingroup$ Then Proof 1 for finite form would work, for $x> \alpha > 1$, where the function is concave $\endgroup$ – Laurent Duval Nov 2 '20 at 17:03

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.