Prove that if $x_1 + x_2 + ... + x_n = n$, then $x_1^k + x_2^k + ... + x_n^k \ge n$ $x_1, x_2, ..., x_n \in \mathbb R$ are nonegative and $k \in \mathbb R$, $k \ge 1$. Prove that if $x_1 + x_2 + ... + x_n = n$, then $x_1^k + x_2^k + ... + x_n^k \ge n$. I tried to find the smallest value of $x_1^k + ... + x_n^k$ and it turned out to be $n$ for $x_1 = ... = x_n = 1$. I did it by proving that if we assume that this expression assumes smallest value for sequence $x_1', x_2', ..., x_n'$ in which not every elements are equal, then we can find a sequence which gives smaller value, by just replacing greatest and smallest $x'$ with their arithmetic mean, which contradicts the "minimality" of this sequence.

Is it a proper proof or does it have some logical faults? I'm not really sure if it is fully correct.
Because by Holder
$$(x_1^k+x_2^k+...+x_n^k)(1+1+...+1)^{k-1}\geq(x_1+x_2+...+x_n)^k,$$
which gives
$$(x_1^k+x_2^k+...+x_n^k)n^{k-1}\geq n^k$$ or
$$x_1^k+x_2^k+...+x_n^k\geq n.$$
Also, we can use Jensen for $f(x)=x^k.$

I think your reasoning is not correct, at lest it's not full for the proof.
Fill in details:

$$n\le \left(x_1+\ldots+x_n\right)^k=x_1^k+\ldots+x_n^k+\left\{\text{non-negative number}\right\}$$

Where did we use that $\;1\le k\in\Bbb R\;$, that the numbers are non-negative, etc ?