Question Number 145240 by loveineq last updated on 03/Jul/21
$$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{cyc}} {\sum}{a}^{\mathrm{3}} +\underset{{cyc}} {\sum}\left({a}+{b}\right)^{\mathrm{3}} \:\leqslant\:\mathrm{27} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} \:\geqslant\:\frac{\mathrm{1}}{\mathrm{2}}\left[{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} \right] \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{For}\:{a}\geqslant{b}\geqslant{c}\geqslant\mathrm{0},\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} \:\leqslant\:\mathrm{2}\left[{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} \right] \\ $$
Answered by mitica last updated on 03/Jul/21
Commented by loveineq last updated on 03/Jul/21
$${thanks}\:{you}.\:{I}\:{find}\:{you}!!!!! \\ $$