What is an upper bound for $\frac{4x - \sigma(x)}{3x - \sigma(x)}$ when $x$ is deficient?

Question

Let $x$ be a positive integer, and let $\sigma(x)$ be the sum of the divisors of $x$. For example, $$\sigma(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28.$$

I would like an upper bound for the expression $$\frac{4x - \sigma(x)}{3x - \sigma(x)},$$ when $x$ is deficient (i.e., when $\sigma(x) < 2x$).

My Attempt

Since $x \leq \sigma(x)$ for all $x$, then under the assumption that $x$ is deficient, we have the bounds $$x \leq \sigma(x) < 2x.$$

Consequently, $-2x < -\sigma(x) \leq -x$, which implies that $$2x < 4x - \sigma(x) \leq 3x$$ and $$x < 3x - \sigma(x) \leq 2x,$$ from which it follows that $$1 < \frac{4x - \sigma(x)}{3x - \sigma(x)} < 3.$$

Question

Is it possible to give an improved (numerical) upper bound of $3$ for $(4x - \sigma(x))/(3x - \sigma(x))$?

Answer 1

Let $x$ be a fixed deficient number. Let $\alpha=\frac{\sigma(x)}{x}$. So $1\le\alpha <2$.

We have $\frac{4x-\sigma(x)}{3x-\sigma(x)}=\frac{4x-\alpha\cdot x}{3x-\alpha\cdot x}=\frac{4-\alpha}{3-\alpha}=1+\frac{1}{3-\alpha}$

For the range of values of $\alpha$, this is bounded above by $1+\frac{1}{3-2}=2$.

This upper bound can be approached, as can be seen by looking at the deficient numbers of the form $x=2^n$, for which $\frac{4x-\sigma(x)}{3x-\sigma(x)}=\frac{2^{n+1}+1}{2^n+1}$

Answer 2

One possible improvement is the following. For all $x\neq 1$ you have that $x+1\le σ(x)$ with equality iff $x$ is prime. So, $$\frac{4x-σ(x)}{3x-σ(x)}\le \frac{3x-1}{3x-σ(x)}<\frac{3x-1}{x}=3-\frac1x$$ For $x=1$, $σ(1)=1$ and hence $$\frac{4-σ(1)}{3-σ(1)}=\frac32<2=3-\frac11$$ so, this bound holds for all deficient $x$'s ($1$ included), that is $$\frac{4x-σ(x)}{3x-σ(x)}<3-\frac1x$$ However this approach gives a bound that depends on $x$.