Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$ 
Problem 1:
Prove that for any $m,n\in\Bbb N_{>0}$, there exists $x \in\Bbb R$ such that
$2\sin n x \cos m x \ge 1$.

Four months ago, someone asked the above question. However, when I wanted to post my answer,
the question was deleted. I searched by Approach0 without any result.
I think that it is a nice question. I don't know why it was deleted. I post it here. I don't remember who posted it before.
Edit 2021/02/20: We may restrict $x$ to be the form $x = r\pi$ where $r$ is a rational number. We give the following problem:

Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that
$2\sin (n r \pi) \cos (m r\pi) \ge 1$.

Any comments and solutions are welcome and appreciated.
Partial results are as follows.
If $n = m$, let $x = \frac{\pi}{4n}$ and we have $2\sin n x \cos m x = \sin 2n x = 1$.
If $n > m$, let $x = \frac{\pi}{2(2n-1)}$. Since $0 < n x < \pi$ and $0 < m x \le (n-1)x < \pi$, we have
\begin{align}
2\sin n x \cos m x &\ge 2\sin n x \cos (n-1)x \\
&= \sin (2n-1)x + \sin x \\
&= 1 + \sin \frac{\pi}{2(2n-1)}\\
& \ge 1.
\end{align}
 A: hint
Using the transformation formula, it will be equivalent to prove that
$$\sin((n+m)x)+\sin((n-m)x)\ge \color{red}{1}$$
Assume that $n\ge m>0$, we can take $$x=\frac{\pi}{2(n+m)}$$
then
$$\sin((n+m)x)=\color{red}{1}$$
and
$$0\le (n-m)x<\frac{\pi}{2}$$
so
$$\sin((n-m)x)\ge\color{red}{0}$$
A: Remarks: Here is my proof for Problem 2. I wrote it for Problem 1 originally, though Problem 1 might be easier. By the way, @mathlove gave a nice proof.
Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that
$2\sin (n r \pi) \cos (m r\pi) \ge 1$.
Proof: Let $x = r\pi$. We split into five cases:

*

*$n > m$: Let $x = \frac{\pi}{2(2n-1)}$. Since $0 < n x < \pi$ and $0 < m x \le (n-1)x < \pi$, we have
\begin{align}
 2\sin n x \cos m x &\ge 2\sin n x \cos (n-1)x \\
 &= \sin (2n-1)x + \sin x \\
 &= 1 + \sin \frac{\pi}{2(2n-1)}\\
 & \ge 1.
\end{align}


*$n = m$: Let $x = \frac{\pi}{4n}$. We have $2\sin n x \cos m x = \sin 2n x = 1$.


*$\frac{5m}{6} < n < m$: Let $x = -\frac{3\pi}{4m}$.
Since $\frac{5\pi}{8} < \frac{3n\pi}{4m}  < \frac{3\pi}{4}$, we have
\begin{align}
 2\sin n x \cos mx &= \sqrt{2}\, \sin \frac{3n\pi}{4m}\\
 &\ge \sqrt{2}\, \sin \frac{3\pi}{4}\\
 & = 1.
\end{align}


*$\frac{m}{6} \le n \le \frac{5m}{6}$: Let $x = -\frac{\pi}{m}$.
Since $\frac{\pi}{6} \le \frac{n\pi}{m} \le \frac{5\pi}{6}$, we have
$$2\sin n x \cos m x = 2\sin \frac{n\pi}{m} \ge 2\sin \frac{\pi}{6} = 1.$$


*$n < \frac{m}{6}$: Let
$$y = \frac{\pi}{m}\Big\lfloor \frac{1}{2}\frac{m}{n} + \frac{1}{2}\Big\rfloor$$
where $\lfloor \cdot \rfloor$ is the floor function.
Clearly, $|\cos my| = 1$. By using $z-1 < \lfloor z \rfloor \le z$ and $\frac{n}{m} < \frac{1}{6}$, we have
$$ny \ge \frac{n\pi}{m}\Big(\frac{1}{2}\frac{m}{n} + \frac{1}{2} - 1\Big)
= \frac{\pi}{2} - \frac{n\pi}{2m} \ge \frac{5\pi}{12}$$
and
$$n y  \le \frac{n\pi}{m}\Big(\frac{1}{2}\frac{m}{n} + \frac{1}{2}\Big)
= \frac{\pi}{2} + \frac{n\pi}{2m} < \frac{7\pi}{12}.$$
Thus, we have $\sin ny \ge \sin \frac{5\pi}{12}$ and thus
$$|2\sin ny \cos my| \ge 2 \sin \frac{5\pi}{12} > 1.$$
If $2\sin ny \cos my > 0$, let $x = y$ and we have $2\sin n x \cos mx > 1$.
If $2\sin ny \cos my < 0$, let $x = -y$ and we have $2\sin n x \cos mx > 1$.
We are done.
