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!