if $\lambda^2\lt2$ then there exist $\epsilon\in\mathbb{Q}, \epsilon\gt0$ such that $(\lambda+\epsilon)^2\lt2$
I've seen it in a proof showing that the set of rational numbers is not complete. It is mentioned also in this answer : https://math.stackexchange.com/q/1612350.
P.S. Since my reputation is below 50 i cannot comment the linked answer to ask them the name of this theorem
As lulu says, the vast majority of mathematical results are not named. We only name results that we refer to a lot. So it has to be something that is not trivial to reproduce and is widely applicable to many situations.
To show this, if $\lambda$ is rational, and assuming $\lambda > 0$ (replace it by its opposite otherwise), you can choose $$\epsilon = \min\left\{\lambda, \dfrac{2-\lambda^2}{4\lambda}\right\}$$
Then $$\epsilon(2\lambda + \epsilon) < \epsilon(4\lambda) \le 2 - \lambda^2$$ $$\lambda^2 + 2\lambda\epsilon + \epsilon^2 < 2\\(\lambda + \epsilon)^2 < 2$$
If $\lambda$ is not rational, then by the Archimedean principle, there is some integer $n > \max\left\{\dfrac 1\lambda, \dfrac{4\lambda}{2-\lambda^2}\right\}$. Setting $\epsilon = \frac 1n$ gives the same inequalities as above.
