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 $N$?
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 $P=2n-\sigma(n)>1$.