Inequality help: $\frac{a+b+c}{3}\geq\sqrt{\frac{ab+bc+ca}{3}}\geq\sqrt[3]{abc}$ I'm working through "Calculus of One Variable" by Joseph Kitchen, and this inequality (problem 3 from section 1.3) is causing me quite a bit of pain to prove:
$$\frac{a+b+c}{3}\geq\sqrt{\frac{ab+bc+ca}{3}}\geq\sqrt[3]{abc},$$
where $a,b,c\geq 0.$
The main issue I'm having is that I cannot use the fact that $0\leq a < b$ if and only if $a^{2} < b^{2}$, as this fact is proven in a later exercise.
If you want to take a look at what is proven and what isn't, the first 80 pages of the book is available on a Google Books preview.
Although a hint for this problem is provided in the back of the book, it has not been useful to me. The hint states: "Use Example 2 of the text. Define $A = (x+y+z)/3$ and $H^{-1} = (x^{-1}+y^{-1}+z^{-1})/3$." Note that $A$ and $H$ refer to the arithmetic and harmonic means, respectively.
Example 2 proved that if $a$, $b$, and $c$ are positive numbers which are not all equal, then $(a+b+c)(bc+ca+ab) > 9abc$. This was proved by noting that $(a+b+c)(bc+ca+ab)-9abc = a(b-c)^{2} + b(c-a)^{2} + c(a-b)^{2}$. With little work, it is not hard to see that when $a$, $b$, and $c$ are non-negative, then $(a+b+c)(bc+ca+ab) \geq 9abc$. However, I'm having a difficult time seeing how this fact is useful for my current problem. Using the hint to define $A = (x+y+z)/3$ and $H^{-1} = (x^{-1}+y^{-1}+z^{-1})/3$, I proved that $A\geq H$:
$$A = \frac{x+y+z}{3}\geq^{(1)} \frac{9xyz}{yz+zx+xy} = \frac{9}{x^{-1}+y^{-1}+z^{-1}}\geq H,$$
where the first inequality (1) holds by Example 2. However, I do not see how this is at all helpful. Here's some of the things that I've tried (that do not involve the hint).
Define $x = \sqrt[3]{a}$, $y = \sqrt[3]{b}$, and $z = \sqrt[3]{c}$. Then, observe that
$$\frac{a+b+c}{3} - \sqrt[3]{abc} \,=\, \frac{x^{3} + y^{3} + z^{3} - 3xyz}{3} = \frac{(x+y+z)\left((x-y)^{2} + (y-z)^{2} + (z-x)^{2}\right)}{6}\geq 0.$$
Thus, $(a+b+c)/3 \geq \sqrt[3]{abc}$. I was also able to prove that $(a+b+c)/3\geq\sqrt{(ab+bc+ca)/3}$, however, I used the fact that $0\leq a < b$ if and only if $a^{2} < b^{2}$, which I'm not supposed to use. I only include this proof since it may give insight into the problem. (Some of these algebraic "tricks" were hard to figure out!)
Since $\frac{1}{2}\left((a-b)^{2} + (b-c)^{2} + (c-a)^{2}\right) = (a+b+c)^{2} - 3(ab + bc + ca)\geq 0$, it follows that $3(a+b+c)^{2}\geq 9(ab+bc+ca)$. And therefore, $(a+b+c)^{2}/9 \geq (ab+bc+ca)/3 \geq 0$. Thus, it follows that $(a+b+c)/3\geq\sqrt{(ab+bc+ca)/3}$.
So... to recap, using the appropriate methods, I have prove that $(a+b+c)/3 \geq \sqrt[3]{abc}$ and that $(a+b+c)/3 \geq3/(a^{-1}+b^{-1}+c^{-1})$. However, the rest is giving me a headache. Perhaps I've missed something obvious, but I'm stuck...

Assuming that you can use $$\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$$ for $x,y,z \ge 0$ Substitute
for $x=ab$, $y=bc$, and $z=ac$.
You will get $$\frac{ab + bc + ac}{3} \ge \sqrt[3]{(abc)^2}$$ and after taking root of both sides you will get the right hand side inequality.
