Ratio of partial sum to total sum

AI Thread Summary
The discussion focuses on proving the inequality that the ratio of the partial sum of the first k numbers to the total sum of n numbers is less than or equal to k/n for non-decreasing sequences. An initial attempt shows that if all numbers are equal, the equality holds. The participant struggles with proving the case when the numbers are not all equal but realizes that rearranging the terms leads to a comparison of means. By defining the mean of the first k numbers and leveraging the properties of non-decreasing sequences, the proof is constructed, confirming that the mean of a subset does not exceed the mean of the entire set. The conclusion affirms the validity of the original inequality.
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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.
 
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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 <br /> = k\overline{X}_k + x_{k+1} + x_{k+2} + \dotsb + x_n<br /> \geq k\overline{X}_k + \overline{X}_k + \overline{X}_k + \dotsb + \overline{X}_k<br /> = 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}
 
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