Question Number 153412 by mathdanisur last updated on 07/Sep/21
$$\mathrm{If}\:\:\:\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in\left(\mathrm{0};\infty\right)\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{bc}}\:+\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{cd}}\:+\:\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{da}}\:+\:\frac{\mathrm{d}^{\mathrm{3}} }{\mathrm{ab}}\:\geqslant\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:+\:\mathrm{d} \\ $$
Answered by puissant last updated on 07/Sep/21
$${a}\geqslant{b}\geqslant{c}\geqslant{d}\:\Rightarrow\:{a}^{\mathrm{3}} \geqslant{b}^{\mathrm{3}} \geqslant{c}^{\mathrm{3}} \geqslant{d}^{\mathrm{3}} \\ $$$${and}\:{then},\:\:\frac{\mathrm{1}}{{bc}}\geqslant\frac{\mathrm{1}}{{cd}}\geqslant\frac{\mathrm{1}}{{da}}\geqslant\frac{\mathrm{1}}{{ab}} \\ $$$${using}\:{rearrangement}\:{inequality},\:{we}\:{get}: \\ $$$${f}\left({a},{b},{c},{d}\right)=\frac{{a}^{\mathrm{3}} }{{bc}}+\frac{{b}^{\mathrm{3}} }{{cd}}+\frac{{c}^{\mathrm{3}} }{{da}}+\frac{{d}^{\mathrm{3}} }{{ab}}\geqslant\frac{{a}^{\mathrm{2}} }{{d}}+\frac{{b}^{\mathrm{2}} }{{a}}+\frac{{c}^{\mathrm{2}} }{{b}}+\frac{{d}^{\mathrm{2}} }{{c}} \\ $$$$\Rightarrow\:{f}\left({a},{b},{c},{d}\right)\geqslant{a}^{\mathrm{2}} ×\frac{\mathrm{1}}{{d}}+{b}^{\mathrm{2}} ×\frac{\mathrm{1}}{{a}}+{c}^{\mathrm{2}} ×\frac{\mathrm{1}}{{b}}+{d}^{\mathrm{2}} ×\frac{\mathrm{1}}{{c}}. \\ $$$${We}\:{have}, \\ $$$${a}\geqslant{b}\geqslant{c}\geqslant{d}\:\Rightarrow\:\frac{\mathrm{1}}{{d}}\geqslant\frac{\mathrm{1}}{{a}}\:\Rightarrow\:\:\frac{{a}^{\mathrm{2}} }{{d}}\geqslant{a} \\ $$$${b}\geqslant{c}\:\Rightarrow\:\frac{\mathrm{1}}{{c}}\geqslant\frac{\mathrm{1}}{{b}}\:\Rightarrow\:\frac{{b}^{\mathrm{2}} }{{c}}+\frac{{c}^{\mathrm{2}} }{{b}}\geqslant\frac{{b}^{\mathrm{2}} }{{b}}+\frac{{c}^{\mathrm{2}} }{{c}}={b}+{c} \\ $$$${c}\geqslant{d}\:\Rightarrow\:\frac{\mathrm{1}}{{d}}\geqslant\frac{\mathrm{1}}{{c}}\:\Rightarrow\:\frac{{c}^{\mathrm{2}} }{{d}}+\frac{{d}^{\mathrm{2}} }{{c}}\geqslant\frac{{c}^{\mathrm{2}} }{{c}}+\frac{{d}^{\mathrm{2}} }{{d}}={c}+{d} \\ $$$${Hence},\:\: \\ $$$$\therefore\because\:\frac{{a}^{\mathrm{3}} }{{bc}}+\frac{{b}^{\mathrm{3}} }{{cd}}+\frac{{c}^{\mathrm{3}} }{{da}}+\frac{{d}^{\mathrm{3}} }{{ab}}\geqslant{a}+{b}+{c}+{d}. \\ $$
Commented by mathdanisur last updated on 07/Sep/21
$$\mathrm{ThankYou}\:\mathrm{Ser} \\ $$