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Question Number 153412 by mathdanisur last updated on 07/Sep/21

$$\mathrm{If}\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in (0;\mathrm{\infty })\mathrm{prove}\mathrm{that}:$$$$\frac{{\mathrm{a}}^3}{\mathrm{bc}}+\frac{{\mathrm{b}}^3}{\mathrm{cd}}+\frac{{\mathrm{c}}^3}{\mathrm{da}}+\frac{{\mathrm{d}}^3}{\mathrm{ab}}⩾\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}$$

Answered by puissant last updated on 07/Sep/21

$$a⩾b⩾c⩾d\Rightarrow a^3⩾b^3⩾c^3⩾d^3$$$$andthen,\frac{1}{bc}⩾\frac{1}{cd}⩾\frac{1}{da}⩾\frac{1}{ab}$$$$usingrearrangementinequality,weget:$$$$f(a,b,c,d)=\frac{a^3}{bc}+\frac{b^3}{cd}+\frac{c^3}{da}+\frac{d^3}{ab}⩾\frac{a^2}{d}+\frac{b^2}{a}+\frac{c^2}{b}+\frac{d^2}{c}$$$$\Rightarrow f(a,b,c,d)⩾a^2\times \frac{1}{d}+b^2\times \frac{1}{a}+c^2\times \frac{1}{b}+d^2\times \frac{1}{c}.$$$$Wehave,$$$$a⩾b⩾c⩾d\Rightarrow \frac{1}{d}⩾\frac{1}{a}\Rightarrow \frac{a^2}{d}⩾a$$$$b⩾c\Rightarrow \frac{1}{c}⩾\frac{1}{b}\Rightarrow \frac{b^2}{c}+\frac{c^2}{b}⩾\frac{b^2}{b}+\frac{c^2}{c}=b+c$$$$c⩾d\Rightarrow \frac{1}{d}⩾\frac{1}{c}\Rightarrow \frac{c^2}{d}+\frac{d^2}{c}⩾\frac{c^2}{c}+\frac{d^2}{d}=c+d$$$$Hence,$$$$\therefore \because \frac{a^3}{bc}+\frac{b^3}{cd}+\frac{c^3}{da}+\frac{d^3}{ab}⩾a+b+c+d.$$

Commented by mathdanisur last updated on 07/Sep/21

$$\mathrm{ThankYou}\mathrm{Ser}$$

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