<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] I think your first method can still do something, but you probably have to use it in some other way, since using Cauchy-Schwarz directly will fail.
This is a suggestion.
Suppose 
$\sqrt{3a+b^3} = u$
$\sqrt{3b+c^3} = v$
$\sqrt{3c+a^3} = w$
Then
$u^2+v^2+w^2 = 3(a+b+c)+a^3+b^3+c^3 = 9 + \sum a^3$
we are going to prove $u+v+w\ge 6$, which is equivalent to prove
$$u^2+v^2+w^2+2uv+2vw+2wu\ge 36$$
That means to prove
$$9 + \sum a^3+2uv+2vw+2wu\ge 36$$
we have proved
$u\ge \dfrac{1}{2}(3\sqrt{a}+\sqrt{b^3})$, thus
$2uv+2vw+2wu \ge \dfrac{1}{2}\sum(9\sqrt{ab}+\sqrt{b^3c^3}+3b^2+3\sqrt{ac^3}) = \\\dfrac{1}{2}\sum( 9\sqrt{ab} +\sqrt{a^3b^3}+3a^2 + 3\sqrt{ab^3})$
We then prove
$$\sum a^3 + \dfrac{1}{2}\sum( 9\sqrt{ab} +\sqrt{a^3b^3}+3a^2 + 3\sqrt{ab^3})\ge 27$$
I checked this with MATLAB, $(1,1,1)$ indeed is a local minimum, but I do not how to deal with the unsymmetric term $\sqrt{ab^3}$, I have tried to use $\sqrt{b^3}\ge 3\sqrt{b}-2$ to make it symmetric, but this will make the inequality fail.
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