Prove that exists such $i$ that $\frac{x_i}{y_i}\geqslant\frac{X}{Y}$

Question

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?

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 $(1)$.