Question Number 144083 by mathdanisur last updated on 21/Jun/21
$${if}\:\:{a};{b};{c}>\mathrm{0}\:\:{then}: \\ $$$$\frac{\mathrm{9}{abc}}{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} }\:+\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:\geqslant\:\mathrm{6} \\ $$
Answered by mitica last updated on 21/Jun/21
$$\frac{\mathrm{9}{abc}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }+\frac{{a}^{\mathrm{2}} }{{bc}}+\frac{{b}^{\mathrm{2}} }{{ac}}+\frac{{c}^{\mathrm{2}} }{{ab}}=\frac{\mathrm{9}{abc}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }+\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}}\:\overset{{am}−{gm}} {\geqslant} \\ $$$$\mathrm{2}\sqrt{\frac{\mathrm{9}{abc}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\centerdot\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}}}=\mathrm{2}\centerdot\mathrm{3}=\mathrm{6} \\ $$
Commented by mathdanisur last updated on 21/Jun/21
$${alot}\:{cool}\:{thanks}\:{Sir} \\ $$