Question Number 144676 by loveineq last updated on 27/Jun/21
$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{{b}}{{ca}}\:\geqslant\:\frac{\mathrm{27}}{\mathrm{8}} \\ $$$$\left(\mathrm{Found}\:\mathrm{by}\:\mathrm{WolframAlpha}\right) \\ $$
Answered by ArielVyny last updated on 27/Jun/21
$$ \\ $$$${suppose}\:{abc}\leqslant\mathrm{1} \\ $$$$\mathrm{27}{abc}\leqslant\mathrm{32} \\ $$$$\frac{\mathrm{27}}{\mathrm{8}}{abc}\leqslant\mathrm{4} \\ $$$$\frac{\mathrm{27}}{\mathrm{8}}\leqslant\frac{\mathrm{4}}{{abc}} \\ $$$${or}\:\mathrm{4}=\left({a}+{b}\right)\left({b}+{c}\right) \\ $$$$\frac{\mathrm{27}}{\mathrm{8}}\leqslant\frac{{cb}+{ac}+{ab}+{b}^{\mathrm{2}} }{{abc}} \\ $$$$\frac{\mathrm{27}}{\mathrm{8}}\leqslant\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{{b}}{{ac}} \\ $$$${then}\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{{b}}{{ac}}\geqslant\frac{\mathrm{27}}{\mathrm{8}} \\ $$$$ \\ $$
Commented by loveineq last updated on 28/Jun/21
$${nice}\:{and}\:{thanks}. \\ $$