<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] By Holder $$\left(\sum_{cyc}\sqrt{3a+b^3}\right)^2\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\geq8(a+b+c)^3=216.$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\leq6$$ or $$\sum_{cyc}\frac{(a+b)^3}{a(a+b+c)^2+3b^3}\leq2$$ or
$$\sum_{cyc}(2a^8b+5a^7b^2-3a^6b^3+3a^6c^3-a^5b^4+17a^5c^4)+$$
$$+abc\sum_{cyc}(12a^6+20a^5b+11a^5c-10a^4b^2-11a^4c^2-3a^4b^3-19a^3b^2c-32a^3c^2b+9a^2b^2c^2)\geq0$$ or
$$1.5\sum_{cyc}(a^8b+a^4b^5-2a^6b^3)+0.5\sum_{cyc}(a^7b^2+a^3b^6-2a^5b^4)+9abc\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)+$$
$$+3abc\sum_{cyc}(a^6-a^3b^3)+2abc\sum_{cyc}(a^5b+a^5c-a^4b^2-a^4c^2)+$$
$$+\sum_{cyc}(0.5a^8b+4.5a^7b^2+2.5a^6c^3+15.5a^5c^4+18a^6b^2c+9a^6c^2b+a^5b^3c-19a^4b^3c^2-32a^4c^3b^2)\geq0,$$
which is true by Schur, Muirhead and AM-GM.
</s>