21. (1) Given that $a, b, c, d, e, f$ are real numbers, and satisfy the following two equations: $a+b+c+d+e+f=10, (a-1)^{2}+(b-1)^{2}+(c-1)^{2}+(d-1)^{2}+(e-1)^{2}+(f-1)^{2}=6$, find the maximum value of $f$. (1993 Balkan Mathematical Olympiad)

(2) Given that $a, b, c, d \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and satisfy the following two equations: $\sin a+\sin b+\sin c+\sin d=1, \cos 2a+\cos 2b+\cos 2c+\cos 2d \geqslant \frac{10}{3}$, prove that $0 \leqslant a, b, c, d \leqslant \frac{\pi}{6}$. (1985 Balkan Mathematical Olympiad)
21. (1) From $a+b+c+d+e+f=10, (a-1)^{2}+(b-1)^{2}+(c-1)^{2}+$ $(d-1)^{2}+(e-1)^{2}+(f-1)^{2}=6$, we get
$$\begin{array}{c}
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}=20 \\
a+b+c+d+e+f=10
\end{array}$$

By the Cauchy-Schwarz inequality, we have
$$5\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right) \geqslant(a+b+c+d+e)^{2}$$

That is, $5\left(20-f^{2}\right) \geqslant(10-f)^{2}$. Solving this, we get $0 \leqslant f \leqslant \frac{10}{3}$. Therefore, the maximum value of $f$ is $\frac{10}{3}$.
(2) Let $x=\sin a, y=\sin b, z=\sin c, u=\sin d$, then $x+y+z+u=1, x^{2}+$ $y^{2}+z^{2}+u^{2} \leqslant \frac{1}{3}$. By the Cauchy-Schwarz inequality, we have $3\left(\frac{1}{3}-u^{2}\right) \geqslant 3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y+$ $z)^{2}=(1-u)^{2}$, that is, $3\left(\frac{1}{3}-u^{2}\right) \geqslant(1-u)^{2}$. Solving this, we get $0 \leqslant u \leqslant \frac{1}{2}$. Since $a, b, c, d \in$ $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we have $0 \leqslant a, b, c, d \leqslant \frac{\pi}{6}$.