Problem
Lets assume we have an inequality such that it involves only positive, real values and all exponents within this inequality are integer multiples of $2n$ (even), we suppose this inequality is true.
Question
If we swap $2n$ with different value, for example $n$, does this preserve validity of this inequality?
Example
Assume $\alpha_i \geq 0, \forall i$ and we assume there are at least two terms. We know following equation is true
$$ \sum_i \alpha_i^2 \leq ( \sum_i \alpha_i )^2 $$
This equation is verified because we can expand $( \sum_i \alpha_i )^2 = \sum_i \alpha_i^2 + \sum_{j \neq i} \alpha_i \alpha_j$,
Does that imply that $\forall n \in \mathbb{N}$ following statement is also true?
$$ \sum_i \alpha_i^{2n} \leq ( \sum_i \alpha_i^n )^{2n} $$
Origin of the problem
I have been trying to show the following for positives $\alpha_i$
$$ (\sum_i \alpha_i^4 )^\frac{1}{4} \leq \sum_i \alpha_i^2 $$
Which is difficult to show algebraically, however if stated property is true it reduces to a trivial form mentioned earlier.
This is false as stated. Here is a concrete counterexample (for the updated version in which there are at least two terms). Given the two terms $\alpha_1 = \alpha_2 = 1/2$, consider when $n=2$.
The LHS of your inequality becomes $(1/2)^4 + (1/2)^4 = 1/8$.
The RHS of your inequality becomes $\big((1/2)^2 + (1/2)^2\big)^4 = (1/2)^4 = 1/16$.
But, $1/8 \not \leq 1/16$.