Question Number 144603 by loveineq last updated on 26/Jun/21
$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}{a}+{b}\right)\left({a}+{b}\right)+\left({b}+\mathrm{2}{c}\right)\left({b}+{c}\right)\:\geqslant\:\mathrm{8}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}+\mathrm{2}{b}+{c}\right)\left({c}+{a}\right)\:\:\:\:\:\:\: \\ $$$$\mathrm{Determine}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}.\:\:\: \\ $$
Answered by mindispower last updated on 27/Jun/21
$$\left(\mathrm{2}{a}+{b}\right)\left({a}+{b}\right)+\left({b}+\mathrm{2}{c}\right)\left({b}+{c}\right) \\ $$$$=\left({a}+{b}\right)^{\mathrm{2}} +\left({b}+{c}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} +{ab}+{bc}+{c}^{\mathrm{2}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left({b}+{c}\right)^{\mathrm{2}} \geqslant\mathrm{2}\left({a}+{b}\right)\left({b}+{c}\right)=\mathrm{8} \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}={b}\left({a}+{c}\right)+{a}^{\mathrm{2}} +{c}^{\mathrm{2}} ..{E} \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\frac{\left({a}+{c}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${E}\Leftrightarrow{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}\geqslant\frac{\left({a}+{c}\right)^{\mathrm{2}} }{\mathrm{2}}+{b}\left({a}+{c}\right)=\left({a}+{c}\right)\left({b}+\frac{{a}+{c}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right)\left(\mathrm{2}{b}+{a}+{c}\right) \\ $$$$\Rightarrow\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right)\left(\mathrm{2}{b}+{a}+{c}\right)+\mathrm{8} \\ $$$$ \\ $$$$ \\ $$
Commented by loveineq last updated on 27/Jun/21
$${thanks} \\ $$
Commented by mindispower last updated on 27/Jun/21
$${pleasur} \\ $$