Question Number 149868 by mathdanisur last updated on 07/Aug/21
$$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{and}\:\:\mathrm{c}\:\:\mathrm{are}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{of}\:\:\mathrm{a} \\ $$$$\mathrm{cuboid}\:\mathrm{with}\:\mathrm{the}\:\mathrm{diagonal}\:\boldsymbol{\mathrm{d}}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{d}\:\leqslant\:\sqrt{\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}}\:+\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{c}}\:+\:\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{a}}} \\ $$
Answered by dumitrel last updated on 08/Aug/21
$$\frac{{a}^{\mathrm{3}} }{{b}}+{ab}\overset{{am}−{gm}} {\geqslant}\mathrm{2}{a}^{\mathrm{2}} \\ $$$$\frac{{b}^{\mathrm{3}} }{{c}}+{cb}\overset{{am}−{gm}} {\geqslant}\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\frac{{c}^{\mathrm{3}} }{{a}}+{ac}\overset{{am}−{gm}} {\geqslant}\mathrm{2}{c}^{\mathrm{2}} \Rightarrow \\ $$$$\Rightarrow\frac{{a}^{\mathrm{3}} }{{b}}+\frac{{b}^{\mathrm{3}} }{{c}}+\frac{{c}^{\mathrm{3}} }{{b}}\geqslant\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({ab}+{bc}+{ac}\right)\geqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={d}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{cool} \\ $$