Question Number 196198 by liuxinnan last updated on 19/Aug/23
$${a}+{b}+{c}+{d}=\mathrm{4} \\ $$$${prove}\:{that}: \\ $$$$\frac{{a}}{{a}^{\mathrm{3}} +\mathrm{8}}+\frac{{b}}{{b}^{\mathrm{3}} +\mathrm{8}}+\frac{{c}}{{c}^{\mathrm{3}} +\mathrm{8}}+\frac{{d}}{{d}^{\mathrm{3}} +\mathrm{8}}\leq\frac{\mathrm{4}}{\mathrm{9}} \\ $$
Answered by AST last updated on 19/Aug/23
$${a}^{\mathrm{3}} +\mathrm{8}={a}^{\mathrm{3}} +\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}\geqslant\mathrm{9}\sqrt[{\mathrm{3}}]{{a}} \\ $$$$\Rightarrow\Sigma\frac{{a}}{{a}^{\mathrm{3}} +\mathrm{8}}\leqslant\frac{\mathrm{1}}{\mathrm{9}}\left(\Sigma{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)\leqslant\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{8}×\frac{{a}+{b}+{c}+{d}}{\mathrm{4}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} =\frac{\mathrm{4}}{\mathrm{9}} \\ $$