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

If a;b;c;d∈(0;∞) prove that: (a^3 /(bc)) + (b^3 /(cd)) + (c^3 /da) + (d^3 /(ab)) ≥ a + b + c + d

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

Answered by puissant last updated on 07/Sep/21

a≥b≥c≥d ⇒ a^3 ≥b^3 ≥c^3 ≥d^3 and then, (1/(bc))≥(1/(cd))≥(1/da)≥(1/(ab)) using rearrangement inequality, we get: f(a,b,c,d)=(a^3 /(bc))+(b^3 /(cd))+(c^3 /da)+(d^3 /(ab))≥(a^2 /d)+(b^2 /a)+(c^2 /b)+(d^2 /c) ⇒ f(a,b,c,d)≥a^2 ×(1/d)+b^2 ×(1/a)+c^2 ×(1/b)+d^2 ×(1/c). We have, a≥b≥c≥d ⇒ (1/d)≥(1/a) ⇒ (a^2 /d)≥a b≥c ⇒ (1/c)≥(1/b) ⇒ (b^2 /c)+(c^2 /b)≥(b^2 /b)+(c^2 /c)=b+c c≥d ⇒ (1/d)≥(1/c) ⇒ (c^2 /d)+(d^2 /c)≥(c^2 /c)+(d^2 /d)=c+d Hence, ∴∵ (a^3 /(bc))+(b^3 /(cd))+(c^3 /da)+(d^3 /(ab))≥a+b+c+d.

$${a}\geqslant{b}\geqslant{c}\geqslant{d}\:\Rightarrow\:{a}^{\mathrm{3}} \geqslant{b}^{\mathrm{3}} \geqslant{c}^{\mathrm{3}} \geqslant{d}^{\mathrm{3}} \\ $$$${and}\:{then},\:\:\frac{\mathrm{1}}{{bc}}\geqslant\frac{\mathrm{1}}{{cd}}\geqslant\frac{\mathrm{1}}{{da}}\geqslant\frac{\mathrm{1}}{{ab}} \\ $$$${using}\:{rearrangement}\:{inequality},\:{we}\:{get}: \\ $$$${f}\left({a},{b},{c},{d}\right)=\frac{{a}^{\mathrm{3}} }{{bc}}+\frac{{b}^{\mathrm{3}} }{{cd}}+\frac{{c}^{\mathrm{3}} }{{da}}+\frac{{d}^{\mathrm{3}} }{{ab}}\geqslant\frac{{a}^{\mathrm{2}} }{{d}}+\frac{{b}^{\mathrm{2}} }{{a}}+\frac{{c}^{\mathrm{2}} }{{b}}+\frac{{d}^{\mathrm{2}} }{{c}} \\ $$$$\Rightarrow\:{f}\left({a},{b},{c},{d}\right)\geqslant{a}^{\mathrm{2}} ×\frac{\mathrm{1}}{{d}}+{b}^{\mathrm{2}} ×\frac{\mathrm{1}}{{a}}+{c}^{\mathrm{2}} ×\frac{\mathrm{1}}{{b}}+{d}^{\mathrm{2}} ×\frac{\mathrm{1}}{{c}}. \\ $$$${We}\:{have}, \\ $$$${a}\geqslant{b}\geqslant{c}\geqslant{d}\:\Rightarrow\:\frac{\mathrm{1}}{{d}}\geqslant\frac{\mathrm{1}}{{a}}\:\Rightarrow\:\:\frac{{a}^{\mathrm{2}} }{{d}}\geqslant{a} \\ $$$${b}\geqslant{c}\:\Rightarrow\:\frac{\mathrm{1}}{{c}}\geqslant\frac{\mathrm{1}}{{b}}\:\Rightarrow\:\frac{{b}^{\mathrm{2}} }{{c}}+\frac{{c}^{\mathrm{2}} }{{b}}\geqslant\frac{{b}^{\mathrm{2}} }{{b}}+\frac{{c}^{\mathrm{2}} }{{c}}={b}+{c} \\ $$$${c}\geqslant{d}\:\Rightarrow\:\frac{\mathrm{1}}{{d}}\geqslant\frac{\mathrm{1}}{{c}}\:\Rightarrow\:\frac{{c}^{\mathrm{2}} }{{d}}+\frac{{d}^{\mathrm{2}} }{{c}}\geqslant\frac{{c}^{\mathrm{2}} }{{c}}+\frac{{d}^{\mathrm{2}} }{{d}}={c}+{d} \\ $$$${Hence},\:\: \\ $$$$\therefore\because\:\frac{{a}^{\mathrm{3}} }{{bc}}+\frac{{b}^{\mathrm{3}} }{{cd}}+\frac{{c}^{\mathrm{3}} }{{da}}+\frac{{d}^{\mathrm{3}} }{{ab}}\geqslant{a}+{b}+{c}+{d}. \\ $$

Commented by mathdanisur last updated on 07/Sep/21

ThankYou Ser

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

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