Suppose I have a sequence of positive real numbers $a_1, \, a_2, \, \dots \,a_n$ such that the following is satisfied:
$\sum \limits_{i=1}^{n}a_i = 1$
I am trying to find the smallest value of $L$ such that the following is also true, for any arbitrary sequence:
$L \geq - \log_n \left( \sum \limits_{i=1}^{n}a_i^{\, 2} \right)$
My question: Is $L$ a constant, or is it best represented as a function of $n$?
By the Cauchy-Schwarz inequality, the condition $\sum a_i = 1$ implies
$$1 = \sum_{i=1}^n 1\cdot a_i \leqslant \sqrt{\sum_{i=1}^n 1^2} \cdot \sqrt{\sum_{i=1}^n a_i^2},$$
so
$$\sum_{i=1}^n a_i^2 \geqslant \frac1n.$$
The bound is sharp, $a_i = \frac1n$ attains it. Therefore
$$-\log_n \sum_{i=1}^n a_i^2 \leqslant - \log_n \frac1n = 1.$$