If $a,b,c\ge 0: ab+bc+ca=3$ then prove $$\sqrt[3]{3a+bc}+\sqrt[3]{3b+ca}+\sqrt[3]{3c+ab}\le \frac{3}{2}\sqrt[3]{a^2+b^2+c^2+29}.$$ I tried to use AM-GM, Cauchy-Schwarz without success.
Notice that $a^2+b^2+c^2+3\ge 2(a+b+c)$ but $$\sqrt[3]{3a+bc}+\sqrt[3]{3b+ca}+\sqrt[3]{3c+ab}\le \frac{3}{2}\sqrt[3]{2(a+b+c)+26}$$ is not true when $a=b=\sqrt{3};c=0.$
I need some help to solve the problem. Thank you for your interest.
By Holder $$\left(\sum_{cyc}\sqrt[3]{3a+bc}\right)^3\leq\sum_{cyc}\frac{3a+bc}{(7a+5b+5c)^2}\left(\sum_{cyc}(7a+5b+5c)\right)^2$$ and it's enough to prove $$\sum_{cyc}\frac{3a+bc}{(7a+5b+5c)^2}\leq\frac{27(a^2+b^2+c^2+29)}{2312(a+b+c)^2}$$ and $uvw$ kills it:
Easy to show that the last inequality it's $f(w^3)\geq0$, where $f$ is a concave function and it's enough to check two following cases.
$abc=0$;
$b=a$.
Can you end it now?
We can get the factor $(7a+5b+5c)^2$ by the following way.
For $k\geq0$ by Holder $$\left(\sum_{cyc}\sqrt[3]{3a+bc}\right)^3\leq\sum_{cyc}\frac{3a+bc}{(ka+b+c)^2}\left(\sum_{cyc}(ka+b+c)\right)^2$$ the equality occurs for $$\frac{\sqrt[3]{3a+bc}}{ka+b+c}=\frac{\sqrt[3]{3b+ac}}{kb+a+c}=\frac{\sqrt[3]{3c+ab}}{kc+a+b}.$$ For $c=0$ and $a=b=\sqrt3$ we have $$\frac{3}{2}\sqrt[3]{a^2+b^2+c^2+29}-\sum_{cyc}\sqrt[3]{3a+bc}=\frac{3}{2}\sqrt[3]{35}-2\sqrt3-\sqrt[3]3=0.000248...$$ Now, from $(a,b,c)=\left(\sqrt3,\sqrt3,0\right)$ and $$\frac{\sqrt[3]{3a+bc}}{ka+b+c}=\frac{\sqrt[3]{3c+ab}}{kc+a+b}$$ we obtain $$k=2\sqrt[6]3-1=1.4018...\approx\frac{7}{5},$$ which gives a legitimation to try to continue with $k=\frac{7}{5}$.
I tried and the inequality turned out true.