From this lecture https://www.udemy.com/course/advanced-calculusreal-analysis-with-the-math-sorcerer/learn/lecture/15077264#overview
It looks like the lecturer made a mistake in the proof. It is hard to Google up this sequence, I was unable to find it anywhere.
I'm assuming $a\in\mathbb{R}$. If it is not the case, please let me know.
By the AM-GM inequality we know that $$\sqrt{1\cdot a^{2}}\leq \frac{1+a^{2}}{2}$$ So $$ \sqrt{a^{2}}<1+a^{2}\implies \left|a\right|<1+a^{2} $$ since $1+a^{2}>0$ then $\left|a\right|<\left|1+a^{2}\right|$, so $1$ is an upper bound on $\frac{\left|a\right|}{\left|1+a^{2}\right|}$