<s>[INST] Prove that $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca$ For all positive $a,b,c $ satisfying $a+b+c = 3$,Prove:
$$
\sum_{cyc} \sqrt[3]{a} \ge \sum_{cyc} ab
$$
This is a hard problem and I tried it myself, but it's really hard without using advanced techniques(e.g. EV theorem or HCF theorem).
Is there any easy way without computer and HCF to this problem?

(Updated) Sorry for that I posted this in a hurry. I will show you more in this updated part.
First of all, HCF kills it.(If you don't know it, click here) Consider
$$
2\sum_{cyc} ab = (\sum_{cyc} a)^2 - \sum_{cyc} a^2 = 9 - \sum_{cyc} a^2
$$
Thus, we only need to prove
$$
\sum_{cyc} (a^2 + 2 \sqrt[3]{a}) \ge 9
$$
or
$$
\sum_{cyc} f(a) \ge 9
$$
where $f(x)=x^2 + 2\sqrt[3]{x}$. And because $f''(x)=2-\frac{4}{9\sqrt[3]{x^5}}$, Thus we know $f(x)$ is convex on $\left [ \frac{\sqrt[5]{8}}{3\sqrt[5]{3}}, 3\right)$. So by HCF theorem we only need to prove when $b=a$,$c=3-2a$ or
$$
6a^2 - 12a + 9 + 4\sqrt[3]{a} +2 \sqrt[3]{3-2a} \ge 9
$$
The rest is easy with derivative.
Second, a friend of mine told me that using tangent line can solve it. But I didn't see how that's relevant.
Third, I found this problem on the Internet, but I'm sure I've solved it or saw the solution to it before.
 [/INST] pqr method:
Using AM-GM, we have
$$\sqrt[3]{a} = \frac{a}{\sqrt[3]{a \cdot a \cdot 1}} \ge \frac{a}{(a + a + 1)/3} = \frac{3a}{2a + 1}.$$
It suffices to prove that
$$\frac{3a}{2a + 1} + \frac{3b}{2b + 1} + \frac{3c}{2c + 1} \ge ab + bc + ca.$$
Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.
It suffices to prove that
$$\frac{36r + 12q + 3p}{8r + 4q + 2p + 1} \ge q$$
or
$$(q + 1)(9 - 4q) + (36 - 8q)r \ge 0.$$
If $9 - 4q \ge 0$, since $q \le p^2/3 = 3$,
the inequality is true.
If $9 - 4q < 0$, using $r \ge \frac{4pq - p^3}{9} = \frac{4}{3}q - 3$ (Schur's inequality), it suffices to prove that
$$(q + 1)(9 - 4q) + (36 - 8q)\cdot \left( \frac{4}{3}q - 3\right) \ge 0$$
or
$$\frac{11}{3}(4q - 9)(3 - q) \ge 0$$
which is true.
We are done.
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