Find all positive real numbers $x$ that satisfy
\[x \sqrt{12 - x} + \sqrt{12x - x^3} \ge 12.\]Enter all solutions, separated by commas.
I think we should start by expanding $\sqrt{12x-x^3}$. That way we get rid of the square roots, but we should be careful because we might introduce some extraneous solutions. We can always check our solutions at the end. Now we can isolate the middle term. So we have \[2x\sqrt{12-x}\sqrt{12x-x^3} = 144 - x^2(12-x) - (12x-x^3).\] That looks a little messy. And we know $x \sqrt{12 - x} \ge 0$. So we can square both sides again. This time we get \[12x - x^3 \ge 144 - 24x \sqrt{12 - x} + x^2(12-x).\] The right side looks a little messy. Let's simplify it. And we can divide both sides by $-12x$. There's a lot of cancellation. So now we can square both sides. We get \[x^2-2(12-x)+\frac{(12-x)^2}{x^2}\le 0.\] That looks a little familiar. We can factor it as \[(x-\frac{12-x}{x})^2\le 0.\]. So we must have $x-\frac{12-x}{x}=0$. So $x^2-12+x=0$. That factors as $(x+4)(x-3)=0$. And the only positive real solutions are $x=3$. So $x=3$ is the only solution to the original inequality.

