Question Number 203891 by York12 last updated on 01/Feb/24
$$\mathrm{Let}\:{a},{b},{c}\:\mathrm{be}\:\mathrm{postive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{a}+{c}}\right)\left(\frac{{b}}{{a}+{c}}+\frac{{c}}{{a}+{b}}\right)\left(\frac{{c}}{{a}+{b}}+\frac{{b}}{{a}+{c}}\right)\geqslant\mathrm{1} \\ $$
Answered by sniper237 last updated on 01/Feb/24
$${The}\:{last}\:{factor}\:{should}\:{be}\:\left(\frac{{c}}{{a}+{b}}+\frac{{a}}{{b}+{c}}\right) \\ $$$${If}\:\:{so}\:,\:{Let}\:{named}\:{P}\:{that}\:{product} \\ $$$${Divide}\:{each}\:{factor}\:{by}\:{c},{a},{b}\:{resp} \\ $$$${P}=\left(\frac{{a}/{c}}{\mathrm{1}+{b}/{c}}+\frac{{b}/{c}}{\mathrm{1}+{a}/{c}}\right).... \\ $$$${P}\geqslant\left(\frac{{a}+{b}}{{c}}\right)\left(\frac{{b}+{c}}{{a}}\right)\left(\frac{{c}+{a}}{{b}}\right)\geqslant\frac{\mathrm{2}\sqrt{{ab}}\mathrm{2}\sqrt{{bc}}\mathrm{2}\sqrt{{ca}}}{{abc}} \\ $$$${Then}\:{P}\geqslant\mathrm{8} \\ $$