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Prove-that-a-b-b-c-c-a-a-2-1-b-2-1-b-2-1-c-2-1-c-2-1-a-2-1-for-a-b-c-are-positive-real-number-




Question Number 150039 by bobhans last updated on 09/Aug/21
Prove that (a/b)+(b/c)+(c/a)≥(√((a^2 +1)/(b^2 +1)))+(√((b^2 +1)/(c^2 +1)))+(√((c^2 +1)/(a^2 +1)))  for a,b,c are positive real number
Provethatab+bc+caa2+1b2+1+b2+1c2+1+c2+1a2+1fora,b,carepositiverealnumber
Answered by EDWIN88 last updated on 09/Aug/21
(•) with Cauchy−Schwarz′s    (a^2 +b^2 )(√((a^2 +1)(b^2 +1))) ≥ (a^2 +b^2 )(ab+1)                                = ab(a^2 +b^2 )+a^2 +b^2                          ≥ ab(a^2 +b^2 +2)     ⇒Σ (a/b)+Σ (b/a) =Σ ((a^2 +b^2 )/(ab))                                      ≥ Σ ((a^2 +b^2 +2)/( (√((a^2 +1)(b^2 +1)))))                        = Σ (√((a^2 +1)/(b^2 +1))) +Σ (√((b^2 +1)/(a^2 +1)))  (•) with Chebyshev′s    Σ (a^2 /b^2 ) = Σ (a^2 /(b^2 +1)) +Σ (a^2 /(b^2 (b^2 +1)))                ≥ Σ (a^2 /(b^2 +1))+Σ (b^2 /(b^2 (b^2 +1)))             = Σ ((a^2 +1)/(b^2 +1))   then (1+Σ (a/b))^2 =1+2(Σ (a/b)+Σ (b/a))+(a^2 /b^2 )          ≥ 1+2(Σ (√((a^2 +1)/(b^2 +1))) +Σ (√((b^2 +1)/(a^2 +1))))+Σ ((a^2 +1)/(b^2 +1))         = (1+Σ (√((a^2 +1)/(b^2 +1))))^2   Thus (a/b)+(b/c)+(c/a) ≥ (√((a^2 +1)/(b^2 +1)))+(√((b^2 +1)/(c^2 +1)))+(√((c^2 +1)/(a^2 +1)))
()withCauchySchwarzs(a2+b2)(a2+1)(b2+1)(a2+b2)(ab+1)=ab(a2+b2)+a2+b2ab(a2+b2+2)Σab+Σba=Σa2+b2abΣa2+b2+2(a2+1)(b2+1)=Σa2+1b2+1+Σb2+1a2+1()withChebyshevsΣa2b2=Σa2b2+1+Σa2b2(b2+1)Σa2b2+1+Σb2b2(b2+1)=Σa2+1b2+1then(1+Σab)2=1+2(Σab+Σba)+a2b21+2(Σa2+1b2+1+Σb2+1a2+1)+Σa2+1b2+1=(1+Σa2+1b2+1)2Thusab+bc+caa2+1b2+1+b2+1c2+1+c2+1a2+1

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