Question Number 145198 by loveineq last updated on 03/Jul/21
$$\mathrm{Let}\:{a}\geqslant{b}\geqslant{c}\geqslant\mathrm{0}\:,\:{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} \neq\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{{a}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} }{{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} }\:\leqslant\:\mathrm{2} \\ $$$$\mathrm{Determine}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}. \\ $$
Answered by mitica last updated on 03/Jul/21
$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\left({u}+{v}\right)\left({u}^{\mathrm{2}} −{uv}+{v}^{\mathrm{2}} \right); \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{\mathrm{6}+{ac}+{bc}−\mathrm{2}{ab}}{\mathrm{3}+\mathrm{2}{ab}−{ac}−{bc}}\leqslant\mathrm{2}\:\bullet \\ $$$${y}=\mathrm{2}{ab}−{ac}−{bc}\Rightarrow{y}=\mathrm{2}{a}\centerdot{b}−\left({a}+{b}\right)\centerdot{c} \\ $$$$\mathrm{2}{a}\geqslant{a}+{b};{b}\geqslant{c}\Rightarrow{y}\geqslant\mathrm{0} \\ $$$${y}\leqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}\left({a}+{b}\right)=\mathrm{3}−{c}^{\mathrm{2}} −{c}\left({a}+{b}\right)=\mathrm{3}−{c}\left({a}+{b}+{c}\right)\leqslant\mathrm{3} \\ $$$$\bullet\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{\mathrm{6}−{y}}{\mathrm{3}+{y}}\leqslant\mathrm{2}\Leftrightarrow\mathrm{0}\leqslant{y}\leqslant\mathrm{3} \\ $$$$ \\ $$
Commented by loveineq last updated on 03/Jul/21
$${thanks} \\ $$