Question Number 103310 by Quvonchbek last updated on 14/Jul/20

Commented byQuvonchbek last updated on 14/Jul/20

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Answered by 1549442205 last updated on 14/Jul/20

Applying Cauchy′s inequality for   three positive numbers we have:  (a^3 /(b+c))+((b+c)/4)+(1/2)≥3 ^3 (√((a^3 /(b+c)).((b+c)/4).(1/2)))=((3a)/2)  Similarly,we have: (b^3 /(c+a))+((c+a)/4)+(1/2)≥((3b)/2)  (c^3 /(a+b))+((a+b)/4)+(1/2)≥((3c)/2).Adding three  above inequalities we get  LHS+((a+b+c)/2)+(3/2)≥((3(a+b+c))/2)  ⇔LHS≥a+b+c−(3/2)(1)  On the other hands,  a+b+c≥3 ^3 (√(abc))=3(2)(due to abc=1   (bythe hypothesis)).From (1) and (2)  we get (a^3 /(b+c))+(b^3 /(c+a))+(c^3 /(a+b))≥(3/2) (q.e.d)  The equality ocurrs if and only if  a=b=c=1  second way:  Applying Cauchy−Schwarz we have:  (a^3 /(b+c))+(b^3 /(c+a))+(c^3 /(a+b))⇔(a^4 /(a(b+c)))+(b^4 /(b(c+a)))+(c^4 /(c(a+b)))  ≥(((a^2 +b^2 +c^2 )^2 )/(2(ab+bc+ca)))≥(((ab+bc+ca)^2 )/(2(ab+bc+ca)))=((ab+bc+ca)/2) (3)  On the other hands,  ab+bc+ca≥3 ^3 (√(ab.bc.ca))=3 ^3 (√((abc)^2 ))=3(4)  (due to abc=1(by the hypothesis))  From(3)and (4) we get  (a^3 /(b+c))+(b^3 /(c+a))+(c^3 /(a+b))≥(3/2).The equality ocurrs  if and only if a=b=c=1(q.e.d)