5. Positive real numbers $a, b, c, d$ satisfy $a+b+c+d=1$. Let $p=\sqrt[3]{7 a+1}+\sqrt[3]{7 b+1}+\sqrt[3]{7 c+1}+\sqrt[3]{7 d+1}$. Then $p$ satisfies ( ).
(A) $p>5$
(B) $p<5$
(C) $p<2$
(D) $p<3$
5.(A).

From $0a^{2}>a^{3}$.
Then $7 a+1=a+3 a+3 a+1$
$$
>a^{3}+3 a^{2}+3 a+1=(a+1)^{3} .
$$

Thus, $\sqrt[3]{7 a+1}>a+1$.
Similarly, $\sqrt[3]{7 b+1}>b+1, \sqrt[3]{7 c+1}>c+1, \sqrt[3]{7 d+1}$ $>d+1$.
Therefore, $p>(a+b+c+d)+4=5$.