Prove that $\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\geq\sqrt[3]{45(ab+ac+bc)}$ 
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
  $$\sqrt[3]{a^2+4bc}+\sqrt[3]{b^2+4ac}+\sqrt[3]{c^2+4ab}\geq\sqrt[3]{45(ab+ac+bc)}$$

A big problem in this inequality there is around $(1,1,0)$.
I tried Holder: 
$$\left(\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\right)^3\sum_{cyc}(a^2+4bc)^3(ka+b+c)^4\geq\left(\sum\limits_{cyc}(a^2+4bc)(ka+b+c)\right)^4$$
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a^2+4bc)(ka+b+c)\right)^4\geq45(ab+ac+bc)\sum_{cyc}(a^2+4bc)^3(ka+b+c)^4,$$
which is nothing for all $k\geq0$.
Of course, we can use Holder with $(ka^2+b^2+c^2+mab+mac+nbc)^4$, but I think in this way even uvw will not help.
I have a proof of the following inequality.

Let $a$, $b$ and $c$ be non-negative numbers and $k=8\cos^340^{\circ}.$ Prove that:
  $$\sqrt[3]{a^2+kbc}+\sqrt[3]{b^2+kac}+\sqrt[3]{c^2+kab}\geq\sqrt[3]{9(1+k)(ab+ac+bc)},$$

but it not so comforts.
Thank you!
 A: Assume that $ab+bc+ca > 0$. Rewrite the inequality as $\sqrt[3]{u} + \sqrt[3]{v} + \sqrt[3]{w} \ge 3$ where
$$u = \frac{27(a^2+4bc)}{45(ab+bc+ca)}, \ v = \frac{27(b^2+4ca)}{45(ab+bc+ca)}, \ w = \frac{27(c^2+4ab)}{45(ab+bc+ca)}.$$
We will use the fact that 
$$\sqrt[3]{x} \ge \frac{3x(5+4x)}{5x^2+20x+2}, \ x \ge 0$$
which follows from 
$$x - \Big( \frac{3x(5+4x)}{5x^2+20x+2}\Big)^3 = \frac{x(125x^2+272x+8)(x-1)^4}{(5x^2+20x+2)^3} \ge 0.$$
Using the fact above, it suffices to prove that
$$\frac{3u(5+4u)}{5u^2+20u+2} + \frac{3v(5+4v)}{5v^2+20v+2} + \frac{3w(5+4w)}{5w^2+20w+2} \ge 3$$
or $f(a,b,c) \ge 0$ where $f(a,b,c)$ is a homogeneous polynomial.
We use the Buffalo Way. WLOG, assume that $c = \min(a,b,c)$. There are two possible cases:
1) $c \le b\le a$: Let $b = c + s, \ a = c+s+t; \ s, t\ge 0$.
Note that $f(c+s+t, c+s, c)$ is a polynomial with non-negative coefficients. True.
2) $c \le a \le b$: Let $a = c + s, \ b = c + s+t; \ s, t \ge 0$.
Note that $f(c+s, c+s+t, c)$ is a polynomial with non-negative coefficients. True.
We are done.
