Let $\sum_{i=1}^n x_i \geqslant X$ and $\sum_{i=1}^n y_i \leqslant Y$ and $X, Y > 0$. Prove that exists such $i$ that $\frac{x_i}{y_i}\geqslant\frac{X}{Y}$.

I think that it could be proved using one of the following ways:

  • pigeonhole principle,
  • induction,
  • by contradiction.

But I got stuck. For example when we try to show it by contradiction we can write that for all $i$ (if $\sum_{i=1}^n y_i$ is positive)

$$\frac{x_i}{y_i} < \frac{X}{Y}$$ $$\sum_{k=1}^n y_k\frac{x_i}{y_i} \leqslant Y\frac{x_i}{y_i} < X \leqslant \sum_{j=1}^n x_j $$ $$\sum_{k=1}^n y_k\frac{x_i}{y_i} < \sum_{j=1}^n x_j $$ $$\frac{x_i}{y_i} < \frac{\sum_{j=1}^n x_j}{\sum_{k=1}^n y_k}$$ But I don't know how it can be useful.

Do you have some hints?


1 Answer 1


We are given that $$ \frac{\sum\limits_{i=1}^nx_i}{\sum\limits_{i=1}^ny_i}\ge\frac XY\tag{1} $$ Suppose that for all $i$, we have $$ \frac{x_i}{y_i}\lt\frac XY\tag{2} $$ Then, multiplying $(2)$ by $y_i$ and summing yields $$ \sum_{i=1}^nx_i\lt\frac XY\sum_{i=1}^ny_i\tag{3} $$ which, after dividing by $\sum\limits_{i=1}^ny_i$, contradicts $(1)$.

  • $\begingroup$ Thank you for this proof. It works at least when $y_i$'s are positive. $\endgroup$ Commented Dec 19, 2015 at 15:07
  • $\begingroup$ I misread the statement that $X\gt0$ and $Y\gt0$ to mean that $x_i\gt0$ and $y_i\gt0$. It is not true if $y_i$'s are not positive. Consider $\frac{3+2}{2+(-1)}\gt\frac42$, but $\frac32\lt\frac42$ and $\frac2{-1}\lt\frac42$ $\endgroup$
    – robjohn
    Commented Dec 19, 2015 at 15:12
  • $\begingroup$ Ok, I think the statement should have more assumptions (about $y_i$'s) in order to be easier provable and not fail as in Stef's example. $\endgroup$ Commented Dec 19, 2015 at 15:31

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