Prove that $3 \le \sum_{cyc}a\sqrt{b^3 + 1} \le \sum_{cyc}ab^2 + 3$ where $a, b, c \ge 0$ and $a + b + c = 3$. 
$a$, $b$ and $c$ are non-negatives such that $a + b + c = 3$. Prove that $$\large 3 \le a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \le \frac{ab^2 + bc^2 + ca^2}{2} + 3$$

This problem is adapted from a competition... that happened two years ago. Yup, I've changed things up.
I have provided a solution below if you want to check out. (I am typing this at midnight so my consciousness is drifting away and I can't focus.)
 A: We have that $a, b, c \ge 0 \iff \sqrt{a^3 + 1} - 1, \sqrt{b^3 + 1} - 1, \sqrt{c^3 + 1} - 1 \ge 0$
$$\implies c\left(\sqrt{a^3 + 1} - 1\right) + a\left(\sqrt{b^3 + 1} - 1\right) + b\left(\sqrt{c^3 + 1} - 1\right) \ge 0$$
$$\iff a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \ge a + b + c = 3$$
The equality sign occurs when $\left[ \begin{align} a = b = 0 &\text{ and } c = 3\\ b = c = 0 &\text{ and } a = 3\\ c = a = 0 &\text{ and } b = 3 \end{align} \right.$.
Furthermore, we have that $$\sqrt{x^3 + 1} = \sqrt{(x^2 - x + 1)(x + 1)} \le \dfrac{(x^2 - x + 1) + (x + 1)}{2} = \dfrac{x^2 + 2}{2}$$
$$\implies a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \le \frac{a(b^2 + 2) + b(c^2 + 2) + c(a^2 + 2)}{2}$$
$$= \frac{ab^2 + bc^2 + ca^2}{2} + (a + b + c) = \frac{ab^2 + bc^2 + ca^2}{2} + 3$$
The equality sign happens when $\left[ \begin{align} a = 0, b = 1 \text{ and } c = 2\\ a = 0, b = 2 \text{ and } c = 1\\ a = 1, b = 2 \text{ and } c = 0\\ a = 1, b = 0 \text{ and } c = 2\\ a = 2, b = 0 \text{ and } c = 1\\ a = 2, b = 1 \text{ and } c = 0\\ \end{align} \right.$.
