Question Number 221262 by Nicholas666 last updated on 28/May/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:{a},\:{b},\:{c},\:>\:\mathrm{0}\:\:,\:\mathrm{show}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{5}} }{{b}^{\mathrm{2}} }\:+\:\frac{{b}}{{c}}\:+\:\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{2}} }\:>\:\mathrm{2}{a} \\ $$$$ \\ $$
Commented by mr W last updated on 31/May/25
$$\frac{{a}^{\mathrm{5}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{c}}+\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{2}} }\geqslant\frac{\mathrm{11}{a}}{\:\sqrt[{\mathrm{11}}]{\mathrm{3}^{\mathrm{9}} ×\mathrm{2}^{\mathrm{8}} }}>\mathrm{2}{a} \\ $$
Answered by breniam last updated on 30/May/25
$$\Leftrightarrow\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{ac}}+\left(\frac{{c}}{{a}}\right)^{\mathrm{3}} >\mathrm{2} \\ $$$$\mathrm{1}°\:{a}^{\mathrm{4}} \leqslant\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\mathrm{1}{A}°\:{b}\:\leqslant\mathrm{2}{ac} \\ $$$${a}^{\mathrm{4}} \leqslant\mathrm{8}{a}^{\mathrm{2}} {c}^{\mathrm{2}} \Rightarrow{c}\geqslant\sqrt[{\mathrm{3}}]{\mathrm{2}}{c}\Rightarrow\left(\frac{{c}}{{a}}\right)^{\mathrm{3}} \geqslant\mathrm{2}\Rightarrow\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{ac}}+\left(\frac{{c}}{{a}}\right)^{\mathrm{3}} >\mathrm{2} \\ $$$$\mathrm{1}{B}°\:{b}>\mathrm{2}{ac} \\ $$$$\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{ac}}+\left(\frac{{c}}{{a}}\right)^{\mathrm{3}} >\mathrm{0}+\mathrm{2}+\mathrm{0}=\mathrm{2} \\ $$$$\mathrm{2}°{a}^{\mathrm{4}} >\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{ac}}+\left(\frac{{c}}{{a}}\right)^{\mathrm{3}} >\mathrm{2}+\mathrm{0}+\mathrm{0}=\mathrm{2} \\ $$
Answered by mr W last updated on 31/May/25
$$\left(\frac{{a}^{\mathrm{5}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{c}}+\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{2}} }\right)\frac{\mathrm{1}}{{a}} \\ $$$$=\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{ca}}+\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} } \\ $$$$=\mathrm{3}×\frac{{a}^{\mathrm{4}} }{\mathrm{3}{b}^{\mathrm{2}} }+\mathrm{6}×\frac{{b}}{\mathrm{6}{ca}}+\mathrm{2}×\frac{{c}^{\mathrm{3}} }{\mathrm{2}{a}^{\mathrm{3}} }\:\:\:\:\:\:\left({tricky}\:{here}!\right) \\ $$$$\geqslant\mathrm{11}\sqrt[{\mathrm{11}}]{\left(\frac{{a}^{\mathrm{4}} }{\mathrm{3}{b}^{\mathrm{2}} }\right)^{\mathrm{3}} \left(\frac{{b}}{\mathrm{6}{ca}}\right)^{\mathrm{6}} \left(\frac{{c}^{\mathrm{3}} }{\mathrm{2}{a}^{\mathrm{3}} }\right)^{\mathrm{2}} }\:\:\:\:\:\:\left({arithmetric}\:{mean}\:\geqslant\:{geometric}\:{mean}\right) \\ $$$$=\mathrm{11}\sqrt[{\mathrm{11}}]{\frac{{a}^{\mathrm{12}−\mathrm{6}−\mathrm{6}} {b}^{−\mathrm{6}+\mathrm{6}} {c}^{−\mathrm{6}+\mathrm{6}} }{\mathrm{3}^{\mathrm{3}} \mathrm{6}^{\mathrm{6}} \mathrm{2}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{11}}{\:\sqrt[{\mathrm{11}}]{\mathrm{3}^{\mathrm{9}} \mathrm{2}^{\mathrm{8}} }}\approx\mathrm{2}.\mathrm{705}>\mathrm{2} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{5}} }{{b}^{\mathrm{2}} }+\frac{{b}}{{c}}+\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{2}} }>\mathrm{2}{a}\:\checkmark \\ $$