<s>[INST] Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$ 
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$


Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
 [/INST] Proof by KaiRain.
By squaring and collecting in terms, the original inequality is equivalent with:
\begin{align*} \sum_{cyc} \sqrt{ \left(3a+b^3 \right) \left(3b+c^3 \right)} \ge \frac{3 \left(a+b \right) \left(b+c \right) \left(c+a \right)}{2} \end{align*}
We will prove that:
\begin{align} \sqrt{ \left(3a+b^3 \right) \left(3b+c^3 \right)} \ge \frac{3ab^2+b^2c}{2}+c^2a+abc \ \ \ (1)\end{align}
holds true for all non-negative reals $a,b,c$ such that $a+b+c=3$. Indeed, this inequality can be proved by C-S and AM-GM in a rather tricky way. Following are the details:
\begin{align*}3= \frac{ \left(a+b+c \right)^2}{3} = ab+bc+ca + \frac{  \left(b+c-2a \right)^2 +3\left(b-c \right)^2 }{12} \ge ab+bc+ca+ \frac{ \left(b-c \right)^2}{4} \end{align*}
Using the above estimation, it turns out:
\begin{align*} \sqrt{ \left(b^3+3a \right) \left(c^3+3b \right)} & \ge \sqrt{ b^3+a \left(ab+bc+ca \right) } \  \sqrt{c^3 + b \left(ab+bc+ca \right) +\frac{b \left(b-c \right)^2}{4}} \\ & = \sqrt{b^3+a^2b+ca^2+abc}  \  \sqrt{ \frac{ b\left(b+c \right)^2}{4}+ab^2+c^3+abc} \\ & \ge  \frac{b^2\left(b+c \right)}{2}+\sqrt{a^3b^3}+c^2a+abc\\ &= \frac{ b^3+\sqrt{a^3b^3}+\sqrt{a^3b^3} }{2} +\frac{b^2c}{2}+c^2a+abc\\ &\ge  \frac{3ab^2}{2} +\frac{b^2c}{2}+c^2a+abc.\end{align*}
Now, by adding two similar inequalities with $(1)$ and taking a note that:
\begin{align*} \sum_{cyc} \left(\frac{3ab^2+b^2c}{2}+c^2a+abc \right) = \frac{3 \left(a+b \right) \left(b+c \right) \left(c+a \right)}{2} \end{align*}
the result follows. Hence, the proof is completed.
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