Does this inequality have any solutions in $\mathbb{N}$?

Question

Does this (number-theoretic) inequality have any solutions $x \in \mathbb{N}$?

$$\frac{\sigma_1(x)}{x} < \frac{2x}{x + 1}$$

Notice that we necessarily have $x > 1$.

Answer 1

Yes; in fact, almost any deficient number will do. Let us show that this is the case for the odd prime numbers $x\in\mathbb{P}$. We know that $\sigma_1(x)=x+1$. On the other hand, the inequality can be rewritten as

$$\sigma_1(x)<\frac{2x^2}{x+1}=2x-2+\frac{2}{x+1}.$$

Since $x+1\le2x-2$ and $\frac{2}{x+1}>0$ for $x\ge3$, the assertion follows.

By the way, I admire your work on odd perfect numbers; it is very educational!