Inequality $\sum_{k=1}^{n} \frac{\log(a_k)}{1+a_{k}^{2}} \leqslant 0$

Question

Let $a_1,a_2,...,a_n$ be positive real numbers such that $a_1 \cdot a_2 \cdot ... \cdot a_n=1$.
Prove that $\sum_{k=1}^{n} \frac{\log(a_k)}{1+a_{k}^{2}} \leqslant 0$.
I tried using Jensen inequality, but function is not concave for all positive real number and I'm stuck.
Any help will be greatly appreciated.

Answer 1

Let $a_k=e^{x_k}.$

Thus, the condition gives $\sum\limits_{k=1}^nx_k=0$ and $$-\sum\limits_{k=1}^n\frac{\ln a_k}{a_k^2+1}=-\sum\limits_{k=1}^n\frac{x_k}{e^{2x_k}+1}=\sum\limits_{k=1}^n\left(\frac{1}{2}x_k-\frac{x_k}{e^{2x_k}+1}\right)=\sum\limits_{k=1}^n\frac{x_k\left(e^{2x_k}-1\right)}{2\left(e^{2x_k}+1\right)}\geq0.$$

Answer 2

We have $$ 2 \sum_{k=1}^{n} \frac{\log(a_k)}{1+a_{k}^{2}} = \sum_{k=1}^{n} \left( \frac{2\log(a_k)}{1+a_{k}^{2}} - \log(a_k)\right) \\ = \sum_{k=1}^{n}\frac{(1-a_k^2)\log(a_k)}{1+a_{k}^{2}} \le 0 $$ since each term in the last sum is $\le 0$.

The inequality is strict unless $a_1 = \cdots = a_n = 1$.

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How came I up with this? First I observed that $$ f(a_1, \ldots, a_k) = \sum_{k=1}^{n} \frac{\log(a_k)}{1+a_{k}^{2}} $$ has the same value if all $a_k$ are replaced by their reciprocals: $$ f(a_1, \ldots, a_k) - f(\frac{1}{a_1}, \ldots, \frac{1}{a_n}) = \sum_{k=1}^{n} \frac{(1+a_k^2)\log(a_k)}{1+a_{k}^{2}} = 0 \, . $$ Then I calculated $$ f(a_1, \ldots, a_k)+f(\frac{1}{a_1}, \ldots, \frac{1}{a_n}) = \sum_{k=1}^{n} \frac{(1-a_k^2)\log(a_k)}{1+a_{k}^{2}} \le 0 \, . $$ These calculations can be “condensed” to the proof given at the beginning.