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

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 $(1)$.
