a.
WLOG assume $a \ge b \ge c \ge d$. Then $\dfrac{a-1}{a} \ge \dfrac{b-1}{b} \ge \dfrac{c-1}{c} \ge \dfrac{d-1}{d}$. By Chebyshev's inequality, we have
$$\sum_{\text{cyc}} \left( a \cdot \frac{a-1}{a} \right) \ge \frac{1}{4} \left( \sum_{\text{cyc}} a \right) \left( \sum_{\text{cyc}} \frac{a-1}{a} \right).$$
Also, by the AM-GM inequality, we have
$$\sum_{\text{cyc}} \frac{a-1}{a} \ge 4 \sqrt[4]{\frac{(a-1)(b-1)(c-1)(d-1)}{abcd}}.$$
Combining these, we obtain
$$(a+b+c+d-4) \ge (a+b+c+d) \sqrt[4]{\frac{(a-1)(b-1)(c-1)(d-1)}{abcd}}.$$
Raising both sides to the fourth power and rearranging the terms, we obtain the desired inequality. Equality holds when $a = b = c = d$.

b.
We have
$$\begin{align*} & \frac{d}{a+b+c} + \frac{a}{b+c+d} + \frac{b}{c+d+a} + \frac{c}{d+a+b} \ge \frac{4}{3} \\ \Leftrightarrow \quad & \frac{a+b+c+d}{a+b+c} + \frac{a+b+c+d}{b+c+d} + \frac{a+b+c+d}{c+d+a} + \frac{a+b+c+d}{d+a+b} \ge \frac{16}{3} \\ \Leftrightarrow \quad & (a+b+c+d) \sum_{\text{cyc}} \frac{1}{a+b+c} \ge \frac{16}{3} \\ \Leftrightarrow \quad & \left( \sum_{\text{cyc}} (a+b+c) \right) \left( \sum_{\text{cyc}} \frac{1}{a+b+c} \right) \ge 16. \end{align*}$$
This is obviously true by the Cauchy-Schwarz inequality. Equality holds when $a = b = c = d$.