Ratio of partial sum to total sum

[txy]

Homework Statement

Given n real numbers $$x_1, x_2, \dotsb , x_n$$ which satisfy $$0 \leq x_1 \leq x_2 \leq \dotsb \leq x_n$$,
show that
$$\frac{x_1 + x_2 + \dotsb + x_k}{x_1 + x_2 + \dotsb + x_n} \leq \frac{k}{n}, \forall 1 \leq k \leq n$$.

Homework Equations

The Attempt at a Solution

If $$x_1 = x_2 = \dotsb = x_n$$, then
$$\frac{x_1 + x_2 + \dotsb + x_k}{x_1 + x_2 + \dotsb + x_n} = \frac{k(x_1)}{n(x_1)} = \frac{k}{n}$$.

If they are not all equal, suppose $$x_1 \neq x_2$$, then everything except $$x_1$$ would be strictly positive. Then I don't know how to continue. I can't seem to get a nice inequality coming out.

Edit: I just realized that by moving the terms around in the above inequality, I get
$$\frac{x_1 + x_2 + \dotsb + x_k}{k} \leq \frac{x_1 + x_2 + \dotsb + x_n}{n}$$
This is like saying that the mean of a set of numbers increases when even bigger numbers are added to the set. This seems intuitive enough, but I haven't figured out how to prove this.

 

[txy]

I think I've solved it.
Let
$$\overline{X}_k = \frac{1}{k}\left(x_1 + x_2 + \dotsb + x_k\right)$$
Then
$$x_1 + x_2 + \dotsb + x_k = k\overline{X}_k$$
Because $$\overline{X}_k$$ is the mean, we have
$$\overline{X}_k \leq x_k \leq x_{k+1} \leq \dotsb \leq x_n$$
So
$$x_1 + x_2 + \dotsb + x_k + x_{k+1} + x_{k+2} + \dotsb + x_n = k\overline{X}_k + x_{k+1} + x_{k+2} + \dotsb + x_n \geq k\overline{X}_k + \overline{X}_k + \overline{X}_k + \dotsb + \overline{X}_k = n\overline{X}_k$$
And therefore
$$\frac{x_1 + x_2 + \dotsb + x_k + x_{k+1} + x_{k+2} + \dotsb + x_n}{n} \geq \overline{X}_k = \frac{x_1 + x_2 + \dotsb + x_k}{k}$$