51 If $a, b, c \in \mathbf{R}^{+}, a+b+c=1$, then the integer part of $M=\sqrt{3 a+1}+\sqrt{3 b+1}+\sqrt{3 c+1}$ is
A. 3
B. 4
C. 5
D. 6
51 B. By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
M & =1 \cdot \sqrt{3 a+1}+1 \cdot \sqrt{3 b+1}+1 \cdot \sqrt{3 c+1} \\
& \leqslant \sqrt{(1+1+1)[(3 a+1)+(3 b+1)+(3 c+1)]} \\
& =3 \sqrt{2} \quad \text{for} \quad a > a^2, \, b > b^2, \, c > c^2, \, \text{so}
\end{aligned}
$$
$$
\begin{aligned}
M & >\sqrt{a^{2}+2 a+1}+\sqrt{b^{2}+2 b+1}+\sqrt{c^{2}+2 c+1} \\
& =(a+1)+(b+1)+(c+1) \\
& =4 .
\end{aligned}
$$

Thus, \(4 < M < 5\).