Notice that if the RHS did not have the term $(a-d)^2$, then
$$\begin{aligned} \sum_{cyc} \frac{a^2}{b} &\geq 4 \\ \Leftrightarrow \sum_{cyc} \left(\frac{a^2}{b} + b\right) &\geq 8 \end{aligned}$$
The above can easily be seen to hold by four applications of the AM-GM inequality. Returning to the original problem, after adding the term $(a-d)^2$ we need to prove
$$\begin{align*} \Leftrightarrow \sum_{cyc} \left(\frac{a^2}{b} + b\right) &\geq 8 + (a-d)^2 \\ \Leftrightarrow \sum_{cyc} \left(\frac{a^2}{b} + b - 2a\right) &\geq (a-d)^2 \\ \Leftrightarrow \sum_{cyc} \left(\frac{a^2 + b^2 - 2ab}{b}\right) &\geq (a-d)^2 \end{align*}$$
By the Cauchy-Schwarz inequality we know
$$\begin{align*} \sum_{cyc} \left(\frac{(a-b)^2}{b}\right) \sum_{cyc} (b) &\geq \left(\sum_{cyc} |a-b|\right)^2 \\ &\geq \left((a-b) + (b-c) + (c-d) + (a-d)\right)^2 = (2a-2d)^2. \end{align*}$$
Therefore
$$\begin{align*} \sum_{cyc} \left( \frac{a^2 + b^2 - 2ab}{b} \right) &\ge \frac{(2a-2d)^2}{a+b+c+d} \\ &= \frac{(2a-2d)^2}{4} = (a-d)^2. \end{align*}$$