# Is the following inequality involving the sum-of-divisors and Euler totient functions true?

First Question

Is the following inequality involving the sum-of-divisors $\sigma$ and Euler totient $\phi$ functions true?

$$\frac{\sigma(N)}{N} \leq \frac{N}{\phi(N)}$$

Second Question

When $N$ satisfies $\sigma(N) = 2N - p$ for some $p > 1$, we say that $N$ is $p$-deficient and we know that

$$\frac{2N}{3N - \sigma(N)} < \frac{\sigma(N)}{N}.$$

Assuming the previous inequality is correct, this gives the upper bound

$$\frac{\phi(N)}{N} < \frac{3}{2} - \frac{\sigma(N)}{N}.$$

(Consider $N = 2^r$ where $r \geq 1$. (Note that $N$ is deficient.) Then $$\frac{\sigma(N)}{N} = \frac{\sigma(2^r)}{2^r} \geq \frac{\sigma(2)}{2} = \frac{3}{2}.$$

This agrees with the fact that $p = 2N - \sigma(N) = 2(2^r) - \sigma(2^r) = 1$.)

These considerations bring me to my next question:

Is there a nice lower bound for $\phi(N)/N$ that will allow us to rule out many, if not most, deficient values for $N$?

• the hint is that $\sigma(N),\phi(N),N$ all are multiplicative, so their product/inverse are too. Then $\sigma(N) / N = \prod_{p | N} 1+\frac{1}{p}$ while $N / \phi(N) = \prod_{p | N} \frac{1}{1-\frac{1}{p}}$ obviously $1+\frac{1}{p} < \sum_{k=0}^\infty \frac{1}{p^k} = \frac{1}{1-\frac{1}{p}}$ hence $\sigma(N) / N < N / \phi(N)$ – reuns Jul 20 '16 at 18:21
• finally $\frac{\sigma(N) \phi(N)}{N^2} = \prod_{p | N} (1+\frac{1}{p})(1-\frac{1}{p}) = \prod_{p | N} 1-\frac{1}{p^2}$ and we get $\frac{1}{\zeta(2)} = \prod_p 1-\frac{1}{p^2} < \frac{\sigma(N) \phi(N)}{N^2} \le 1$ – reuns Jul 20 '16 at 18:26

I just found the following closely related MSE post.

The answer in that post gives the following inequality: $$\frac{6}{{\pi}^2} < \frac{\sigma(n)\phi(n)}{n^2} \leq 1.$$

The upper bound so given answers my first question.

The lower bound gives: $$\frac{6}{{\pi}^2}\cdot\frac{n}{\sigma(n)} < \frac{\phi(n)}{n}.$$

If I let $n$ satisfy $\sigma(n) = 2n - P$ for some $P > 1$, then $$\frac{6}{{\pi}^2}\cdot\frac{n}{\sigma(n)} < \frac{\phi(n)}{n} < \frac{3}{2} - \frac{1}{2}\cdot\frac{\sigma(n)}{n}.$$

Letting $\sigma(n)/n = I(n)$, I get $$\frac{6}{{\pi}^2} < \frac{3}{2}I(n) - \frac{1}{2}\cdot\left(I(n)\right)^2 = \frac{I(n)}{2}\cdot\left(3 - I(n)\right).$$

Setting $x=I(n)$, according to WolframAlpha this last inequality is equivalent to $$0.48307 \approx \frac{3}{2} - \frac{\sqrt{3\left(3{\pi}^2 - 16\right)}}{2\pi} < x < \frac{3}{2} + \frac{\sqrt{3\left(3{\pi}^2 - 16\right)}}{2\pi} \approx 2.51693,$$ which are trivial bounds for $x=I(n)$, when $P=2n-\sigma(n)>1$.

• 3 names for $\sigma(n)/n = I(n) = x$... – reuns Jul 20 '16 at 21:00