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.
 A: I think, TL method does not help here.
Another way.
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, $x^3+y^3+z^3=3$ and we need to prove that:
$$x+y+z\geq x^3y^3+x^3z^3+y^3z^3$$ or
$$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=u^3(9u^3-9uv^2+w^3)^5-(9v^6-9uv^2w^3+w^6)^3.$$
But, $$f'(w^3)=5u^3(9u^3-9uv^2+w^3)^4+2(9uv^2-2w^3)(9v^6-9uv^2w^3+w^6)^2\geq0,$$ which says that $f$ increases and it's enough to prove our inequality for the minimal value of $w^3$.
Now, $x$, $y$ and $z$ are positive roots of the equation
$$(t-x)(t-y)(t-z)=0$$ or
$$t^3-3ut^2+3v^2t-w^3=0,$$ which says that a graph of $f(t)=t^3-3ut^2+3v^2t$ and the line $g(t)=w^3$ have three common points.
But $$f'(t)=3t^2-6ut+3v^2=3(t-(u+\sqrt{u^2-v^2}))(t-(u-\sqrt{u^2-v^2})),$$
which gives $t_{min}=u+\sqrt{u^2-v^2}$ and since $f(0)=0$, we obtain two cases:

*

*$f\left(t_{min}\right)>0$.

In this case $w^3$ gets a minimal value, when the line $g(t)=w^3$ is a tangent line to the graph of $f$, which says that in this case it's enough to prove our inequality for equality case of two variables.
We'll prove this for the inequality $$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$
It's enough to assume $y=z=1$ and we need to prove that:
$$(x+2)^3(x^3+2)^5\geq243(2x^3+1)^3$$ or $h(x)\geq0,$ where
$$h(x)=3\ln(x+2)+5\ln(x^3+2)-3\ln(2x^3+1)-5\ln3.$$
But $$h'(x)=\frac{(x-1)(3x^5+7x^4+7x^3+6x^2-x-1)}{(x+2)(x^3+2)(2x^3+1)},$$
which gives $x_{min}=1$, $x_{max}=0.375...$ and since $h(1)=0$ it's enough to prove that $h(0)\geq0$ or $256\geq243,$ which is true.


*$w^3\rightarrow0^+$.

Let $z\rightarrow0^+$ and $y=1$.
Thus, we need to prove that $$(x+1)^3(x^3+1)^5\geq243x^9,$$ which is true by AM-GM because $256>243.$
A: 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.
A: It's easy by considering using Hölder inequality, which can remove roots effectively.
First we homogenise it, we need to prove
$$
(\sum_{cyc}\sqrt[3]{a})^3(\sum_{cyc} a )^5 \ge 3^5(\sum_{cyc}ab)^3
$$
By Hölder we have
$$
(\sum_{cyc}\sqrt[3]{a})^3(\sum_{cyc} a )^5 \ge (\sum_{cyc}\sqrt[4]{a^3})^8
$$
So, we only need to prove
$$
(\sum_{cyc}\sqrt[4]{a^3})^8 \ge 243 (\sum_{cyc}ab)^3
$$
This is the well-known 243-inequality (by Harazi on AOPS).
$$
(\sum_{cyc} x^3)^8 \ge 243(\sum_{cyc} x^4y^4)^3
$$
(In case you didn't notice, this can be proved directly from A-G inequality)
