Show $x^n \geq x_1^n+x_2^n+\ldots+x_k^n \Bigg\vert x_1+x_2+\ldots+x_k = x, x \geq 0, n \geq 1, k\in\mathbb{Z}$ Hello StackExchange Community,
This is my first post on the forum. Please forgive me for any errors with formatting and my expressions. I am working on the following proof:
Show $$x^n \geq x_1^n+x_2^n+\ldots+x_k^n \Bigg\vert x_1+x_2+\ldots+x_k = x, x \geq 0, n \geq 1, k\in\mathbb{Z}$$
This is to say, I'd like to show that a positive number raised to $n\geq1$ is greater than the sum of any set of addends each raised to $n$.
What I have scribbled so far:

Let's look at $x^n \geq (\frac{x}{2})^n+(\frac{x}{2})^n$
$x^n \geq2 \frac{x^n}{2^n}$
$x^n \geq 0$
$1 \geq \frac{2}{2^n}$
$2^n \geq 2$
$n \geq 1$, thus true.
It is obvious that $x^n \geq k(\frac{x^n}{k^n})$ is also true (for $k$
  $x_k$ that are all equal).
Can I make $x_1^n+x_2^n+\ldots+x_k^n$ look like $k(\frac{x}{k})^n$?
  Well if $x_1^n+x_2^n+\ldots+x_k^n < k(\frac{x}{k})^n$, then it is also
  less than $x^n$ (proving original problem).
$k \geq 0$
$\frac{k^n}{k}(x_1^n+x_2^n+\ldots+x_k^n) < x^n$
I feel I need to use the information $x_1+x_2+\ldots+x_k = x$
$\frac{k^n}{k}(x_1^n+x_2^n+\ldots+x_k^n) < (x_1+x_2+\ldots+x_k)^n$

I am not sure how to revise my thoughts and continue. I greatly appreciate any comments and questions; thank you in advance for posting! :)
With regards,
Polite Master
 A: Unfortunately your idea won't lead to a solution, because evenly distributing will give the least possible sum. Notice that the single $n$-th power is in fact still a sum of $k\ $ $n$-th powers of numbers that have the same sum, and so essentially the problem is about maximizing the sum of the $n$-th powers given the sum of $k$ non-negative numbers. One simple approach would be to see that if you can prove that $a^n + b^n \le (a+b)^n$ for any non-negative real $a,b$, then then you can use it $(k-1)$ times to prove the result. Do you see it?
A: Take any rear number  $0\le y\le1.$ Then $$1=(y+(1-y))^n\ge y^n+(1-y)^n$$ is true for any $n\in\mathbb{N}.$
Take any $x_1,x_2\ge 0$ and put $y=\dfrac{x_1}{x_1+x_2}$ $$\implies(x_1+x_2)^n\ge x_1^n+x_2^n$$ Now induct on $k.$
A: You want to show that $(x_1+x_2+\cdots+x_k)^n \ge x_1^n+x_2^n+\cdots+x_k^n$ whenever each $x_i$ is $\ge 0$. But this is obvious if you expand the left-hand side: you will get all the terms $x_1^n,x_2^n,\ldots,x_k^n$ plus some non-negative terms.
With this in mind, you should be able to prove the proposition more formally, using induction on $k$.
