Question Number 84792 by M±th+et£s last updated on 16/Mar/20
$${a},{b},{c}\geqslant\mathrm{0} \\ $$$${a}+{b}+{c}=\mathrm{3} \\ $$$${show}\:{that} \\ $$$$\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:+\sqrt[{\mathrm{3}}]{{c}}\geqslant{ab}+{bc}+{ca} \\ $$