15. (14th "Hope Cup" Mathematics Invitational Test) Let the function $f(x)=\sqrt[3]{1+x}-\lambda x$, where $\lambda>0$.
(1) Find the range of $\lambda$ such that $f(x)$ is a monotonic function on the interval $[0,+\infty)$;
(2) Can this monotonicity be extended to the entire domain $(-\infty,+\infty)$?
(3) Solve the inequality $2 x-\sqrt[3]{1+x}<12$.
15. (1) Let $x_{1}>x_{2} \geqslant 0$, then $f\left(x_{1}\right)-f\left(x_{2}\right)=\sqrt[3]{1+x_{1}}-\sqrt[3]{1+x_{2}}-\lambda\left(x_{1}-x_{2}\right)=\left(x_{1}-\right.$ $\left.x_{2}\right)\left[\frac{1}{\sqrt[3]{\left(1+x_{1}\right)^{2}}+\sqrt[3]{\left(1+x_{1}\right)\left(1+x_{2}\right)}+\sqrt[3]{\left(1+x_{2}\right)^{2}}}-\lambda\right]$. Clearly, $\sqrt[3]{\left(1+x_{1}\right)^{2}}+\sqrt[3]{\left(1+x_{1}\right)\left(1+x_{2}\right)}$ $+\sqrt[3]{\left(1+x_{2}\right)^{2}}>3$. Therefore, when $\lambda \geqslant \frac{1}{3}$, $f\left(x_{1}\right)-f\left(x_{2}\right)<0$,
which proves that $f(x)$ is a decreasing function on $[0,+\infty)$.
It is easy to verify that the range $\lambda \geqslant \frac{1}{3}$ is the largest possible.
(2) The decreasing interval $[0,+\infty)$ cannot be extended further, which can be proven by contradiction, but this is omitted here.
(3) From (1) and (2), we know that the function $u=2 x-\sqrt[3]{1+x}$ is monotonically increasing on $(0,+\infty)$. Also, $u(7)=12$, so the solution set of the original inequality is $(-\infty, 7)$.