13) $a$ and $b$ are two numbers greater than or equal to zero. We know that: $a^{3}+a<b-b^{3}$. What is the correct order among the three numbers $a, b$ and 1?
(A) $b<a<1$
(B) $a=b=1$
(C) $a<1<b$
(D) $a<b<1$
(E) $1<a<b$.
13. The answer is $(\mathbf{D})$.

Since $a \geq 0$, we have $a^{3}+a \geq 0$ and therefore from the inequality in the problem it follows that $b-b^{3}>0$. On the other hand, $b-b^{3}=b\left(1-b^{2}\right)$ and for this quantity to be strictly positive (considering that $b \geq 0$), both $b>0$ and $b<1$ must hold. Therefore, $0 < b < 1$.

Since $b-b^{3}>0$ (as proven above), it must be that $b-b^{3}<b$. Therefore, from the inequality $a+a^{3} \leq b-b^{3}$ it follows that $a<b$. We have thus proven $a<b<1$. [Problem proposed by X.Y. Lu.]