10. (2006 National High School Mathematics League Zhejiang Preliminary Contest Question) $\max _{a, b, \cdots \in \mathbf{R}^{+}} \min \left\{\frac{1}{a}, \frac{1}{b^{2}}, \frac{1}{c^{3}}, a+b^{2}+c^{3}\right\}=$
10. Solution Let $t=\min \left\{\frac{1}{a}, \frac{1}{b^{2}}, \frac{1}{c^{3}}, a+b^{2}+c^{3}\right\}$, then $0<t \leqslant \frac{1}{a}, 0<t \leqslant \frac{1}{b^{2}}, 0<t \leqslant \frac{1}{c^{3}}$, which implies $a \leqslant \frac{1}{t}, b^{2} \leqslant \frac{1}{t}, c^{3} \leqslant \frac{1}{t}$. Therefore, we have $t \leqslant a+b^{2}+c^{3} \leqslant \frac{3}{t}$. Thus, we can get $t \leqslant \sqrt{3}$, and when $a=b^{2}=c^{3}=\frac{\sqrt{3}}{3}$, $t=\sqrt{3}$. Therefore, $\max _{4,6} \min \left\{\frac{1}{a}, \frac{1}{b^{2}}, \frac{1}{c^{3}}, a+b^{2}+c^{3}\right\}=\sqrt{3}$.