Question
Let $\alpha _1,\alpha _2\in (0,1)$ be real numbers. Prove that for a given positive integer $N$, there are $n,m_1,m_2\in \mathbb{Z}, n\leq N$ such that for $i\in \{1,2\},|\alpha _i-\frac{m_i}{n}|\leq \frac{1}{n\sqrt(N)}$
Attempt
The bound in this question is different from that of Dirichlet’s theorem which requires that $|\alpha -\frac{m}{n}|\leq \frac{1}{nN}$, and I don’t seem to know how to construct this bound. I tried to approach the proof this way:
Consider $N+1$ points in the plane ${(x_i,y_{1i},y_{2i})}$, where $x_i=\frac{i}{N}$ for $i=0,1,…,N$, and $y_{ji}=\alpha _jx_i$ (for $j=1,2$). So, each $x_i$ has two corresponding $y_{ji}$ values on two lines, $y=\alpha _1x$ and $y=\alpha _2x$.
Now, let’s divide $(0,1)$ into $N$ equal sub-intervals. Since we have $N+1$ points, then by the pigeonhole principle, at least two points must fall into the same sub-interval along the $x$-axis, say $x_p$ and $x_q$, where $p<q$. This means $x_q-x_p$ and $y_{jq}-y_{jp}$ (for $j=1,2$) corresponds to the difference between the coordinates of these two points. Now, we have $0<x_q-x_p<\frac{1}{N}$ i.e. $0<N(x_q-x_p)<1$. Let $n=N(x_q-x_p)$ and $m_j=N(y_{jq}-y_{jp})$. Since $y_{ji}=\alpha _jx_i$, it follows that $\alpha _jn-m_j$ is small for $j=1,2$, specifically smaller than $\frac{1}{n\sqrt(N)}$.
Note: I know constructing the bound may significantly affect the way I would approach the proof. Your help would be greatly appreciated.