Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$

I would start with noticing that $a^{2} + 1 \geq 2\sqrt{a^{2}} = 2a$.

Similarly, we have that $b^{2} + 1 \geq 2b$.

Then we have that \begin{align*} \frac{a^{2} + 1}{b} + \frac{b^{2} + 1}{a} \geq 2\left(\frac{a}{b} + \frac{b}{a}\right) \geq 4\sqrt{\frac{a}{b}\times\frac{b}{a}} = 4 \end{align*}

Hopefully this helps!

One can actually just use AM-GM directly: $$\frac{a^2}b+\frac1b+\frac{b^2}a+\frac1a\geq 4\sqrt[4]{\frac{a^2}{b}\cdot\frac{1}{b}\cdot\frac{b^2}a\cdot\frac1a}=4\sqrt[4]{1}=4.$$

I assume that $a,b$ are positive real numbers. Then we may assume $a\geq b$, hence $a^2+1\geq b^2+1$ and $\frac{1}{a}\leq\frac{1}{b}$, so the rearrangement inequality gives $$\frac{a^2+1}{b}+\frac{b^2+1}{a}\geq\frac{a^2+1}{a}+\frac{b^2+1}{b}\geq4$$ where the last inequality holds by what you already showed.