10. If $f(x)=\sin x$, (1) If $f(x)+1 \geq a x+\cos x$ holds for all $x \in [0, \pi]$, find the range of values for $a$
(2) Prove: $\sum_{k=1}^{n+1} f\left(\frac{k \pi}{2 n+1}\right) \geq \frac{3 \sqrt{2}(n+1)}{4(2 n+1)}$
10. Analysis: (1) According to the problem, $\sin x + 1 \geq a x + \cos x$ holds for all $x \in [0, \pi]$. Let $g(x) = \sin x + 1 - a x - \cos x$.
$g'(x) = \cos x - a + \sin x = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right) - a$,
and since $x \in [0, \pi]$,
we have $\sqrt{2} \sin \left(x + \frac{\pi}{4}\right) \in [-1, \sqrt{2}]$.

When $a \leq -1$, $g'(x) \geq 0$ always holds, so $g(x) \geq g(0) = 0$, and thus $a \leq -1$ is valid.
When $a \geq \sqrt{2}$, $g'(x) \leq 0$ always holds, so $g(x) \geq g(\pi) = 2 - a \pi \geq 0$, which implies $a \leq \frac{2}{\pi}$, a contradiction!
When $1 < a < \sqrt{2}$, there exists $x_0 \in (0, \pi)$ such that $g(x)$ is increasing for $x \in (0, x_0)$ and decreasing for $x \in (x_0, \pi)$. Therefore, $g_{\min}(x) = \{g(0), g(\pi)\}$. We only need $g(0) \geq 0$ and $g(\pi) \geq 0$, which gives $-1 \leq a \leq \frac{2}{\pi}$.
In summary, the range of $a$ is $a \leq \frac{2}{\pi}$.
(2) From (1), for $x \in [0, \pi]$, $\sin x - \cos x \geq \frac{2}{\pi} x - 1$, so $\sqrt{2} \sin \left(x - \frac{\pi}{4}\right) \geq \frac{2}{\pi} x - 1$.
When $1 \leq k \leq n+1$, $0 \leq \frac{k \pi}{2 n + 1} + \frac{\pi}{4} \leq \pi$, thus
$$
\sqrt{2} \sin \frac{k \pi}{2 n + 1} = \sqrt{2} \sin \left(\frac{k \pi}{2 n + 1} + \frac{\pi}{4} - \frac{\pi}{4}\right) \geq \frac{2}{\pi} \left(\frac{k \pi}{2 n + 1} + \frac{\pi}{4}\right) - 1 = \frac{2 k}{2 n + 1} - \frac{1}{2}
$$

Therefore, $\sqrt{2} \sum_{k=1}^{n+1} f\left(\frac{k \pi}{2 n + 1}\right) \geq \sum_{k=1}^{n+1} \left(\frac{2 k}{2 n + 1} - \frac{1}{2}\right) = \frac{(n+1)(n+2)}{2 n + 1} - \frac{n+1}{2} = \frac{3(n+1)}{2(2 n + 1)}$,
hence, $\sum_{k=1}^{n+1} f\left(\frac{k \pi}{2 n + 1}\right) \geq \frac{3 \sqrt{2} (n+1)}{4 (2 n + 1)}$.