Rounding to nearest integer in the log domain

Question

This is about the threshold of whether to round down to $N$ or round up to $N+1$ where the proportional rounding error (or the error in the log domain) is smallest.

What is the simplest way to prove

$$\begin{align} \log(N) + \log\big(1 + \tfrac{1}{2N} \big) &= \\ \\ \log \big( N + \tfrac12 \big) &> \tfrac12 \big( \log(N) + \log(N+1) \big) \\ \\ &= \log(N) + \tfrac12 \log \big( 1 + \tfrac1N \big) \\ \end{align}$$

or that

$$ \log\big(1 + \tfrac{1}{2N} \big) > \tfrac12 \log \big( 1 + \tfrac1N \big) $$

for any positive integer $N \in \mathbb{Z} > 0$ ?

I can sorta do it with L'Hopital's rule. But there should be a simpler way.

Answer 1

Start from the inequality

$$ 1 + {1 \over N} + {1 \over 4N^2} > 1 + {1 \over N}. $$

Factorize the left hand side to get

$$ \left( 1 + {1 \over 2N} \right)^2 > {1 + {1 \over N}}. $$

Take the square root of both sides to get

$$ 1 + {1 \over 2N} > \sqrt{1 + {1 \over N}} $$

and take logs.

Alternately, this follows from the well known AM-GM inequality applied to $1$ and $1 + 1/N$.