Question Number 220987 by fantastic last updated on 21/May/25
$${Let}\:{a},{b},{c}\:{be}\:{positive}\:{reals}\:{such}\:{that}\:{abc}=\mathrm{1}.{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} \left({b}+{c}\right)}+\frac{\mathrm{1}}{{b}^{\mathrm{3}} \left({c}+{a}\right)}+\frac{\mathrm{1}}{{c}^{\mathrm{3}} \left({a}+{b}\right)}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 21/May/25
$$=\frac{{abc}}{{a}^{\mathrm{3}} \left({b}+{c}\right)}+\frac{{abc}}{{b}^{\mathrm{3}} \left({c}+{a}\right)}+\frac{{abc}}{{c}^{\mathrm{3}} \left({a}+{b}\right)} \\ $$$$=\frac{\left(\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}}+\frac{\left(\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{a}}}+\frac{\left(\frac{\mathrm{1}}{{c}}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} \\ $$$$\geqslant\frac{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)^{\mathrm{2}} }{\mathrm{2}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)}=\frac{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}}{\mathrm{2}} \\ $$$$\geqslant\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{abc}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$