Ratio of partial sum to total sum

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SUMMARY

The discussion centers on proving the inequality that the ratio of the partial sum of the first k real numbers to the total sum of n real numbers is less than or equal to the ratio of k to n, specifically expressed as \frac{x_1 + x_2 + \dotsb + x_k}{x_1 + x_2 + \dotsb + x_n} \leq \frac{k}{n}. The proof utilizes the concept of means, demonstrating that the mean of the first k numbers is less than or equal to the last number in the ordered sequence, leading to the conclusion that adding larger numbers increases the overall mean. The final proof is established by manipulating the sums and applying the properties of means.

PREREQUISITES
  • Understanding of inequalities in mathematics
  • Familiarity with the concept of means and averages
  • Basic knowledge of real number sequences
  • Ability to manipulate algebraic expressions
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  • Study the properties of inequalities in real analysis
  • Learn about the Cauchy-Schwarz inequality and its applications
  • Explore the concept of weighted averages and their implications
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Students studying real analysis, mathematicians interested in inequalities, and educators looking for examples of mean-related proofs.

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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.
 
Last edited:
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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}
 
Last edited:

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