Zeros of $f(\alpha) = \sum_{i=1}^r p_i^\alpha - \sum_{i=1}^t q_i^\alpha$ (Apostol ANT 4.17)

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I'm working through Introduction to Analytic Number Theory by Tom Apostol. I've come across this question (Chapter 4, Exercise 17),

Given an integer $n > 1$ with two factorizations $n = \prod_{i=1}^r p_i$ and $n = \prod_{i=1}^t q_i$, where the $p_i$ are primes (not necessarily distinct) and the $q_i$ are arbitrary integers $> 1$. Let $\alpha$ be a nonnegative real number.

(a) If $\alpha \geq 1$ prove that

$$\sum_{i=1}^r p_i^\alpha \leq \sum_{i=1}^t q_i^\alpha.$$

(b) Obtain a corresponding inequality relating these sums if $0 \leq \alpha < 1$.

I've been able to do part (a), (essentially relating sums to products), but I can't come up with anything for part (b). I've understood that, if

$$f(\alpha) = \sum_{i=1}^r p_i^\alpha - \sum_{i=1}^t q_i^\alpha$$

then $f$ is monotonic and $f(\alpha_0) = 0$ for some $\alpha_0$ in $0 \leq \alpha_0 < 1$ that is dependent on $n$ and the factorization of $n$ into $q_i$.

I've seen this Math SE post, which pretty much asks the same question, except there hasn't been an answer yet after a little more than a year.

Also in this solution manual (page 60), the author Greg Hurst acknowledges that the closed form for $\alpha_0$ is difficult to find and provides some experimental data for minimum values of $\alpha_0$ (pages 61-62).

So the ultimate question: is there a closed form for $\alpha_0$, or some pattern to $\alpha_0$? Is there some relationship between these sums when $0 \leq \alpha < 1$ or is establishing one similar to the inequality of part (a) impossible?

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$$xy\ge x+y\iff (x-1)(y-1)\ge 1$$

Thus with $n=\prod_i p_i $ where $ p_i\le p_{i+1}$ and $\alpha \ge 0$ then $$\sum_i p_i^\alpha=\inf\{\sum_j d_j^\alpha, \prod_j d_j=n,d_j\in \Bbb{Z}_{\ge 2}\}$$ implies $\forall i<l, (p_i^\alpha-1)(p_l^\alpha-1)\ge 1$ which implies $$(p_2^\alpha-1)(p_1^\alpha-1)\ge 1$$

Conversely if $(p_2^\alpha-1)(p_1^\alpha-1)\ge 1$ then for all $uv| n$ we have $(u^\alpha-1)(v^\alpha-1)\ge 1$ thus $u^\alpha v^\alpha \ge u^\alpha+v^\alpha$ and hence the minimizer is $\sum_i p_i^\alpha$.

Whence $\sum_i p_i^\alpha$ is the minimizer iff $\alpha\ge\alpha_0$ where $\alpha_0$ is the largest real zero of $(p_2^a-1)(p_1^a-1)-1$