(10) Given $a>0, b>0, a^{3}+b^{3}=1$, then the range of values for $a+b$ is
Solution: $(1, \sqrt[3]{4}]$
From $0<a b \leqslant \frac{(a+b)^{2}}{4}$ and $1=a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)=(a+b)\left[(a+b)^{2}-3 a b\right]$
we have $\frac{1}{4}(a+b)^{3} \leqslant 1<(a+b)^{3}$, so $1<a+b \leqslant \sqrt[3]{4}$.