Let $L > 1 $ be an integer such that $$ L= p_{1}p_{2}...p_{n} = q_{1}q_2 q_3 ... q_m$$ where $p_i$ are primes which are not necessarily distinct and $q_i$ are arbitrary integers.
If I take $L=100$, then $p_1=p_2=2$ and $p_3 = p_4 = 5$ and let $q_1 = q_2 = 2$ and $ q_3 = 25$, then we observe that $$ \sum_{1}^{4}p_i^x \leq \sum_{1}^{3}q_i^x $$ for any $x \geq 1$
Also if I take many $L$ and test for different $x$ the inequality $$ \sum_{1}^{n}p_i^x \leq \sum_{1}^{m}q_i^x $$ for any $x \geq 1$ seems to be obvious and very intuitive but proving it seems to be monstrous task.
Is the above inequality true for any $x$ greater than or equal to 1 or does it holds when $x$ is between $0$ and $1$ or does there exist a $L$ such that the above condition is not satisfied?
The above inequality is true for any $x \ge 1$. We only need the following lemma:
$$\forall a,b \in \mathbb N\setminus\{1\}: a^x+b^x\le (ab)^x$$
Which is proven by observing:
$$\frac 1{a^x}+\frac1{b^x}\le \frac1a+\frac1b\le\frac12+\frac12=1$$
and multiplying the result by $(ab)^x$ on both sides. This proves the inequality after strong induction on $n$, i.e. the number of prime divisors (including multiplicities), as follows:
The case $n=1$ is trivial and the case $n=2$ follows from the lemma.
Suppose the result holds for $n=k$. Consider $L = \prod_{i=1}^{k+1} p_i$.
-Suppose $q_i \ne L\ \forall i$, and choose $q_j \ne 1$.
Let $S$ be a subset of $\{1, \dots, n\}$ such that $\prod_{i \in S} p_i = q_j$.
By induction hypothesis, $\sum_{i \in S}p_i^x \le q_j^x$ and $\sum_{i \notin S} p_i^x \le \sum_{i\ne j} q_i^x$.
Hence $\sum_{i=1}^n p_i^x \le q_j^x+\sum_{i\ne j} q_i^x =\sum_{i=1}^m q_i^x$.
-The case $q_j=L$ follows easily from induction as well:
$$\sum_{i=1}^m q_i^x \ge L^x \ge p_1^x + \left(\frac L {p_1}\right)^x \ge p_1^x + p_2^x + \left(\frac L {p_1p_2}\right)^x \ge \dots \ge \sum_{i=1}^n p_i^x$$
This concludes the induction step.
The inequality is not necessarily true for $0<x<1$. For $n>m$, when $x \to 0$:
$$\sum_{i=1}^n p_i^x\to n, \sum_{i=1}^m q_i^x\to m$$
hence eventually the inequality fails. For example, we have $2^{1/2} + 2^{1/2} > 4^{1/2}$.