Question Number 176174 by Matica last updated on 14/Sep/22
$$ \\ $$$$\:\mathrm{a},{b},{c}\:\in\mathbb{R}_{+\:} ^{\ast} \:\:\mathrm{and}\:{abc}=\mathrm{1}.\:\mathrm{Prove} \\ $$$$\:\:\:\frac{{c}\:\sqrt{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:+\:\frac{{b}\:\sqrt{{a}^{\mathrm{3}} +{c}^{\mathrm{3}} }}{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }\:+\:\frac{{a}\sqrt{{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$
Answered by behi834171 last updated on 14/Sep/22
$$\Sigma\frac{{c}\sqrt{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{{abc}.\Pi\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)}{\Pi\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}}\geqslant \\ $$$$\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\sqrt{{a}^{\mathrm{3}} {b}^{\mathrm{3}} }×\mathrm{2}\sqrt{{b}^{\mathrm{3}} {c}^{\mathrm{3}} }×\mathrm{2}\sqrt{{c}^{\mathrm{3}} {a}^{\mathrm{3}} }}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }×\mathrm{2}\sqrt{{b}^{\mathrm{2}} {c}^{\mathrm{2}} }×\mathrm{2}\sqrt{{c}^{\mathrm{2}} {a}^{\mathrm{2}} }}}\geqslant \\ $$$$\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}=\mathrm{3}\geqslant\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\:\:.\blacksquare \\ $$