<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] From the original inequality $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
Let’s set $$x= \sqrt{3a + b^3}$$ $$y= \sqrt{3b + c^3}$$ $$z= \sqrt{3c + a^3}$$
By AM-GM we know that $$x^2-2xy+y^2 \ge 0$$ $$y^2-2yz+z^2 \ge0$$ $$z^2-2xz+x^2 \ge 0$$ So we then can set up the following inequalities
$$x^2+y^2\ge2xy$$
$$y^2+z^2\ge2yz$$
$$z^2+x^2\ge2xz$$
Multiply each inequality above by $z$, $x$, $y$ in that order which yields
$$zx^2+zy^2\ge2xyz$$
$$xy^2+xz^2\ge2xyz$$
$$yz^2+yx^2\ge2xyz$$
Add these inequalities and factor to get
$$xy(x+y)+xz(x+z)+zy(y+z) \ge 6xyz$$
Now divide both sides by $xyz$ for
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\ge6$$
Now replace $x$, $y$ and $z$ with the terms on the LHS of the original inequality above
$$\frac{\sqrt{3a + b^3}+\sqrt{3b + c^3}}{\sqrt{3c + a^3}}+\frac{\sqrt{3a + b^3}+\sqrt{3c + a^3}}{\sqrt{3b+ c^3}}+ \frac{\sqrt{3b + c^3}+\sqrt{3c + a^3}}{\sqrt{3a+ b^3}} \ge 6$$
with equality if and only if $a=b=c=1$.
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