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If-a-b-c-d-0-prove-that-a-3-bc-b-3-cd-c-3-da-d-3-ab-a-b-c-d-




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
Ifa;b;c;d(0;)provethat:a3bc+b3cd+c3da+d3aba+b+c+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.
abcda3b3c3d3andthen,1bc1cd1da1abusingrearrangementinequality,weget:f(a,b,c,d)=a3bc+b3cd+c3da+d3aba2d+b2a+c2b+d2cf(a,b,c,d)a2×1d+b2×1a+c2×1b+d2×1c.Wehave,abcd1d1aa2dabc1c1bb2c+c2bb2b+c2c=b+ccd1d1cc2d+d2cc2c+d2d=c+dHence,∴∵a3bc+b3cd+c3da+d3aba+b+c+d.
Commented by mathdanisur last updated on 07/Sep/21
ThankYou Ser
ThankYouSer

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