Prove that there exists infinitely many positive integers $n$ such that $\sin^2{(na)}+\sin^2{(nb)}\le \frac{2\pi^2}{n}$

Here is my idea (if I'm wrong, I will delete my answer):

According to the Dirichlet's theorem for any given irrational $\alpha$ there are infinitely many integers $p,q$ such that $$|\alpha - \frac{p}{q}|<\frac{1}{q^2} $$

Let's take $\alpha = a/\pi, a \in \mathbb{R}; \ q = n, p = k$ $$|an-\pi k|<\frac{\pi}{n}$$ Thus there are infinitely many positive integers $n, k$ such that $$|\sin(an)|=|\sin(an)-\sin( \pi k)| < \frac{\pi}{n}<\frac{\pi}{\sqrt{n}}$$

Thus inequality $\sin^2(an) < \frac{\pi^2}{n}$ is strict.

EDIT: There is a problem when $a \neq b$ then the sets of $n$'s for them may not intersect, so I was searching for some generalizations of Dirichlet's theorem and it seems I found one (Corrolary 1B at Ch2, paragraph 1 in Schmidt - Diophantine approximation):

For any given irrational $\alpha_1, \alpha_2$, the system of inequalities $$|\alpha_j - \frac{p_j}{q}|<\frac{1}{q\sqrt{q}}, \ j=1,2 $$ has infinite number of solutions $p_1,p_2,q \in \mathbb{Z}$.

Then by taking (irrational) $\alpha_1=a/\pi, \alpha_2 = b/\pi$, and repeating steps above we see that there exist infinite number integer solutions $n,k_1,k_2$ for which $$|an - \pi k_1|<\frac{\pi}{\sqrt{n}}, \ |bn - \pi k_2|<\frac{\pi}{\sqrt{n}} $$ which gives us desired inequality. So the last thing left is to prove it when $a,b$ are such that $\alpha_1,\alpha_2$ are rational.

In this case $a=\pi\frac{p}{q}$, so $\sin(na)=0$ for $n=qk, \ k \in \mathbb{Z}$; the same holds for $b$ with another coefficient: $n=lk$; Thus $n=qlk, \ k \in \mathbb{Z}$ will do it.

Using the identity $|\sin x| < |x|$ we can try to check a "linearized" version of the inequality:

$$ ||n a||_{\mathbb{R}/\mathbb{Z}}^2 + ||n b ||_{\mathbb{R}/\mathbb{Z}}^2 < \frac{2\pi^2}{n}$$

where $ ||x||_{\mathbb{R}/\mathbb{Z}} = \min \big\{ |x - n\pi | : n \in \mathbb{Z}\big\}$. Both terms are positive; maybe we can get both error terms to be small:

$$ ||n a||_{\mathbb{R}/\mathbb{Z}} < \frac{\pi}{\sqrt{n}} \hspace{0.25in}\text{and}\hspace{0.25in} ||n b ||_{\mathbb{R}/\mathbb{Z}} < \frac{\pi}{\sqrt{n}} $$

Dirichlet's approximation in $\mathbb{R}^2$ says we can do just that. Why only $\frac{1}{\sqrt{n}}$ ? This is a bigger error than the 1D Dirichlet theorem.

We can prove Dirichlet's theorem using the pigeonhole principle. Consider the square $[0,\pi]^2$ and divide it into $\sqrt{N} \times \sqrt{N}$ pieces

$$ [0, \pi]\times [0, \pi] =\bigcup_{0 \leq m,n < \sqrt{N}} \Big[ \tfrac{(m-1/2)\pi}{\sqrt{N}}, \tfrac{(m+1/2)\pi}{\sqrt{N}}\Big]\times \Big[ \tfrac{(n-1/2)\pi}{\sqrt{N}}, \tfrac{(n+1/2)\pi}{\sqrt{N}}\Big]$$

By pigeonhole, one square must contain at least $2$ - in fact infinitely many - of the values $(na, nb) \in \mathbb{R}^2 / \mathbb{Z}^2$ with $n \in \mathbb{Z}$. $$ \Big((m-n)a, (m-n)b\Big) \in \Big[ -\tfrac{\pi}{\sqrt{N}}, \tfrac{\pi}{\sqrt{N}}\Big]^2 $$

For more, see What is your favorite application of the Pigeonhole Principle?

For example $n = 100, a = \sqrt{2}, b = \sqrt{3}$. Try to get an error within $\frac{1}{10}$.

A classical problem in simultaneous approximation.

Let $a_n=(\lambda n\pmod{\pi},\mu n\pmod{\pi})\in\mathbb{T}$ and let we fix some huge $N$.

Let $Q_n\subset\mathbb{T}$ be a square centered in $a_n$ having side length $\frac{\pi}{\sqrt{N}}$.

Among $Q_1,\ldots,Q_N$, at least two squares have to overlap, since the sum of their Lebesgue measures equals $\mu(\mathbb{T})$. The linearity of the sequence $\{a_n\}_{n\geq 1}$ hence implies that there is some $M\leq N$ (for clarity, $M$ is the absolute difference of the indices of the previous overlapping squares) such that both $\lambda M\pmod{\pi}$ and $\mu M\pmod{\pi}$ are $\leq\frac{\pi}{\sqrt{N}}$.

The claim follows by considering that $\sin x$ is a Lipschitz function with Lipschitz constant $1$.