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.
Real Analysis – Proving the Inequality Sum of log(a_k)/(1+a_k^2) ? 0
inequalitymultivariable-calculusreal-analysistangent-line-method
Best Answer
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 $a_1 = \cdots = a_n = 1$.
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.