$$
\begin{array}{l}
\text { 2. Given } a>b>c>1, M=a-\sqrt{c}, \\
N=a-\sqrt{b}, P=2\left(\frac{a+b}{2}-\sqrt{a b}\right), \\
Q=3\left(\frac{a+b+c}{3}-\sqrt[3]{a b c}\right) .
\end{array}
$$

Then the smallest among $M, N, P, Q$ is $(\quad)$.
2.C.

Obviously, $M>N$.
Since $Q-P=c+2 \sqrt{a b}-3 \sqrt[3]{a b c}$
$$
>3 \sqrt[3]{c \cdot \sqrt{a b} \cdot \sqrt{a b}}-3 \sqrt[3]{a b c}=0 \text {, }
$$

Therefore, $Q>P$.
$$
\begin{array}{l}
\text { Also, } P-N=b-2 \sqrt{a b}+\sqrt{b} \\
=\sqrt{b}[(\sqrt{b}-\sqrt{a})+(1-\sqrt{a})]<0,
\end{array}
$$

Hence, $P<N$.
Thus, $P$ is the smallest.