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Given-a-b-c-nonnegative-numbers-which-satisfy-a-b-c-3-Prove-that-1-2ab-2-1-1-2bc-2-1-1-2ac-2-1-1-

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Question Number 157598 by naka3546 last updated on 25/Oct/21
Given a,b,c nonnegative numbers which satisfy a+b+c = 3. Prove that (1/(2ab^2 + 1)) + (1/(2bc^2 +1)) + (1/(2ac^2 + 1)) ≥ 1 .
Givena,b,cnonnegativenumberswhichsatisfya+b+c=3.Provethat12ab2+1+12bc2+1+12ac2+1⩾1.
Answered by ghimisi last updated on 25/Oct/21
((abc))^(1/3) ≤((a+b+c)/3)⇒abc≤1 Σ(1/(2ab^2 +1))=Σ((((1/b))^2 )/(2a+(1/b^2 )))≥((((1/a)+(1/b)+(1/c))^2 )/(2(a+b+c)+(1/a^2 )+(1/b^2 )+(1/c^2 ))) ((((1/a)+(1/b)+(1/c))^2 )/(6+(1/a^2 )+(1/b^2 )+(1/c^2 )))≥^((•)) 1 (•)⇔(1/a^2 )+(1/b^2 )+(1/c^2 )+2((1/(ab))+(1/(bc))+(1/(ac)))≥6+(1/a^2 )+(1/b^2 )+(1/c^2 )⇔ a+b+c≥3abc⇔abc≤1
abc3⩽a+b+c3⇒abc⩽1Σ12ab2+1=Σ(1b)22a+1b2⩾(1a+1b+1c)22(a+b+c)+1a2+1b2+1c2(1a+1b+1c)26+1a2+1b2+1c2⩾(∙)1(∙)⇔1a2+1b2+1c2+2(1ab+1bc+1ac)⩾6+1a2+1b2+1c2⇔a+b+c⩾3abc⇔abc⩽1
Commented by naka3546 last updated on 25/Oct/21
Thank you, sir.
Thankyou,sir.

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