Question Number 144634 by loveineq last updated on 27/Jun/21
$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\frac{\mathrm{2}{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{{c}+{a}}\:\geqslant\:\mathrm{6} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +\mathrm{3}{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\frac{\mathrm{3}{b}\left({c}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)}{{c}+{a}}\:\geqslant\:\mathrm{8} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\frac{\mathrm{4}{b}\left({c}^{\mathrm{4}} +{a}^{\mathrm{4}} \right)}{{c}+{a}}\:\geqslant\:\mathrm{10} \\ $$
Answered by mindispower last updated on 27/Jun/21
$${c}^{\mathrm{2}} +{a}^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({c}+{a}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\geqslant{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{b}\left({c}+{a}\right) \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ac}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\geqslant\left({b}^{\mathrm{2}} +{bc}+{ac}+{cb}\right)+{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$${b}^{\mathrm{2}} +{bc}+{ac}+{cb}+\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}={t} \\ $$$$\left({b}+{a}\right)\left({b}+{c}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\geqslant\mathrm{4}+\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{2}{ab}+\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{2}{bc}+\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} +\frac{\mathrm{2}{ac}}{\mathrm{4}}=\mathrm{4}+\frac{\mathrm{1}}{\mathrm{2}}\left({b}^{\mathrm{2}} +{ac}+{bc}+{ab}\right) \\ $$$$=\mathrm{4}+\frac{\mathrm{1}}{\mathrm{2}}\left({b}+{a}\right)\left({b}+{c}\right)=\mathrm{6} \\ $$$$\Leftrightarrow\geqslant\mathrm{6} \\ $$$$ \\ $$$$ \\ $$
Answered by mindispower last updated on 27/Jun/21
$$\left(\mathrm{2}\right)\:{i}\:{think}\:\frac{\mathrm{3}{b}\left({c}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)}{{c}+{a}}….? \\ $$$$\left(\mathrm{3}\right)…\left(\mathrm{4}\right)\:{Weighted}\:{AM}\:{worcks} \\ $$$$ \\ $$
Commented by loveineq last updated on 27/Jun/21
$${yes},\:{is}\:\frac{{c}^{\mathrm{3}} +{a}^{\mathrm{3}} }{{c}+{a}}. \\ $$