Approximation of logarithm of harmonic mean

Question

Given a large $M\in\mathbb{N}$,if I could approximate $M\log\left(1+\frac{1}{\sum_{i=1}^M x_i^{-1}}\right)$ by $\sum_{i=1}^{M}\log(1+\frac{x_i}{M})$ in the case of $x_i\in\mathbb{R}$, $0<x\leq C $ ($C$ is a constant predefined) ?

Answer 1

This would be an odd thing to do since, when every $x_i\to0$ (which respects the constraint that $0\lt x_i\leqslant C$ for every $i$, for some fixed $C$), the LHS is equivalent to the harmonic mean of $(x_i)_{1\leqslant i\leqslant M}$ and the RHS is equivalent to its arithmetic mean, that is, $$ \left(\frac1M\sum_{i=1}^M\frac1{x_i}\right)^{-1}\quad\text{vs}\quad\frac1M\sum_{i=1}^Mx_i. $$