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 $a_i$? $\endgroup$ Nov 1, 2020 at 21:05

1 Answer 1


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. $\endgroup$ Nov 2, 2020 at 2:58
  • $\begingroup$ Then Proof 1 for finite form would work, for $x> \alpha > 1$, where the function is concave $\endgroup$ Nov 2, 2020 at 17:03

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