First note that $\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c}\;$ form a triangle. Indeed,

$\begin{align}(\sqrt[3]{a}+\sqrt[3]{b})^3&=a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})\\ &\gt a+ b\\ &\gt c, \end{align}$

implying $\sqrt[3]{}+\sqrt[3]{b}\gt\sqrt[3]{c}.\;$ With this in mind, denote

$-\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=2x,\\ \sqrt[3]{a}-\sqrt[3]{b}+\sqrt[3]{c}=2y,\\ \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}=2z.$

Then $x,y,z\gt 0.\;$ Further $\sqrt[3]{a}=y+z,\;$ $\sqrt[3]{b}=x+z,\;$ and $\sqrt[3]{c}=x+y.\;$ The required inequality becomes

$4(x^3+y^3+z^3)\ge\sqrt[3]{3}(x+y+z)-1.$

Let $x+y+z=3s.\;$ By Jensen's inequality, $x^3+y^3+z^3\ge 3s^3,\;$ with equality only if $x=y=z=s.\;$ Hence, suffice it to show that

$12s^3-3\sqrt[3]{3}s+1\ge 0,$

which is equivalent to $(2\sqrt[3]{3}s-1)^2(\sqrt[3]{3}s+1)\ge 0,\;$ which is obviously true.

Equality holds only if $\displaystyle x=y=z=\frac{1}{2\sqrt[3]{3}},\;$ i.e., when $\displaystyle a=b=c=\frac{1}{3}.$

By the AM-GM inequality,

$\displaystyle\begin{align} (\sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c})^3+\frac{2}{3}&=(\sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c})^3+\frac{1}{3}+\frac{1}{3}\\ &\ge 3\sqrt[3]{(\sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c})^3\cdot\frac{1}{3}\cdot\frac{1}{3}}\\ &=\frac{3}{\sqrt[3]{9}}(\sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c})\\ &=\sqrt[3]{3}(\sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}). \end{align}$

That is,

$\displaystyle(\sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c})^3+\frac{2}{3}\ge\sqrt[3]{3a}+\sqrt[3]{3b}-\sqrt[3]{3c}.$

Similarly,

$\displaystyle(\sqrt[3]{b}+\sqrt[3]{c}-\sqrt[3]{a})^3+\frac{2}{3}\ge\sqrt[3]{3b}+\sqrt[3]{3c}-\sqrt[3]{3a}$

and

$\displaystyle(\sqrt[3]{c}+\sqrt[3]{a}-\sqrt[3]{b})^3+\frac{2}{3}\ge\sqrt[3]{3c}+\sqrt[3]{3a}-\sqrt[3]{3b}$

Adding the three gives the required inequality. Equality is attained for $\displaystyle a=b=c=\frac{1}{3}.$

I am grateful to Dan Sitaru for communicating to me the above problem form his book Math Accent, with two solutions. Solution 1 is by Le Giugiuc, Solution 2 is by Dan Sitaru.