If $a,b,c>0: a+b+c=3$ then$$\sqrt[3]{\frac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3}}\ge \frac{\sqrt[3]{abc}+2}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}.$$
I can prove the following inequality$$\sqrt[3]{\frac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3}}\ge \frac{3\sqrt[3]{abc}}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}.$$It is weaker than the starting one because $\sqrt[3]{abc}+2\ge 3\sqrt[3]{abc}\iff 1\ge \sqrt[3]{abc}.$
We may use Cauchy-Schwarz $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}=3$$and by AM-GM $$\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\ge 3\sqrt[9]{a^2b^2c^2}\ge 3\sqrt[3]{abc}.$$
But I can not use same idea for the starting inequality.
Can you help me? Thank you.
Update.
I also tried AM-GM $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge \frac{3}{\sqrt[3]{abc}}$$
Thus, we need to prove
$$\frac{1}{\sqrt[9]{abc}}\ge \frac{\sqrt[3]{abc}+2}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}.$$
But I don't know how to continue the proof.
Best Answer
We have $$ \begin{aligned} &\sqrt[3]{ab}\sqrt[3]{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}}=\sqrt[3]{\frac{a+b+\frac{ab}{c}}{3}}\\ ={}&\frac1{3}\sqrt[3]{(a+b+c)(a+b+c)\Big(a+b+\frac{ab}{c}\Big)}\geqslant\frac1{3}(a+b+\sqrt[3]{abc}). \end{aligned} $$ Hence $$ \begin{aligned} &(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac})\sqrt[3]{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}}\geqslant\frac23(a+b+c)+\sqrt[3]{abc}\\ \iff{}& \begin{aligned} \sqrt[3]{\frac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3}}\ge \frac{\sqrt[3]{abc}+2}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}} \end{aligned}. \end{aligned} $$