# Lower bound of weighted average of sequence

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 $$a_i, \quad\text{with}\quad i= 1,2,\ldots,N$$?

• Unless I went wrong, the inequality does not seem valid everywhere. Don't you have more constraint on the $a_i$? Nov 1, 2020 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. 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: 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.

• Thank you very much for your help. You are right. The assumption is ai >> 1 to make this lower bound valid. Nov 2, 2020 at 2:58
• Then Proof 1 for finite form would work, for $x> \alpha > 1$, where the function is concave Nov 2, 2020 at 17:03