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Question Number 194808 by York12 last updated on 15/Jul/23

suppose a,b,c are positive real numbers prove the inequality (((a+b)/2))(((b+c)/2))(((c+a)/2))≥(((a+b+c)/3))(((abc)^2 ))^(1/3)

supposea,b,carepositiverealnumbersprovetheinequality(a+b2)(b+c2)(c+a2)⩾(a+b+c3)(abc)23

Commented by York12 last updated on 16/Jul/23

guys , please help

guys,pleasehelp

Commented by sniper237 last updated on 16/Jul/23

call each member L and R Use arihmeico−geomeric inegaligy n=3 , ^3 (√((abc)^2 )) =^3 (√((ab)(ac)(bc))) ≤ ((ab+ac+bc)/3) recall (a+b)(a+c)(b+c)=(a+b+c)(ab+ac+bc)+abc L−R≥(((a+b)(a+c)(b+c)−8abc)/(8×9))≥0 cause for n=2, 2(√(ab)) ≤a+b ; 2(√(ac)) ≤a+c ; 2(√(bc)) ≤b+c

calleachmemberLandRUsearihmeico−geomericinegaligyn=3,3(abc)2=3(ab)(ac)(bc)⩽ab+ac+bc3recall(a+b)(a+c)(b+c)=(a+b+c)(ab+ac+bc)+abcL−R⩾(a+b)(a+c)(b+c)−8abc8×9⩾0causeforn=2,2ab⩽a+b;2ac⩽a+c;2bc⩽b+c

Commented by York12 last updated on 16/Jul/23

thanks sir

thankssir

Commented by York12 last updated on 17/Jul/23

(((a+b)/2))(((b+c)/2))(((a+c)/2))−(((a+b+c)/3))(((abc)^2 ))^(1/3) ≥ (((a+b)/2))(((b+c)/2))(((a+c)/2))_((((a+b+c)(ab+bc+ac))/9)) =(((a+b)(b+c)(a+c))/8)−(((a+b)(b+c)(a+c)+abc)/9) =(a+b)(b+c)(a+c)((1/8)−(1/9))−((abc)/9)≥((abc)/9)−((abc)/9)=0 ⇒(((a+b)/2))(((b+c)/2))(((a+c)/2))−(((a+b+c)/3))(((abc)^2 ))^(1/3) ≥0 ⇒(((a+b)/2))(((b+c)/2))(((a+c)/2))≥(((a+b+c)/3))(((abc)^2 ))^(1/3) →(Hence proved)

(a+b2)(b+c2)(a+c2)−(a+b+c3)(abc)23⩾(a+b2)(b+c2)(a+c2)_((a+b+c)(ab+bc+ac)9)=(a+b)(b+c)(a+c)8−(a+b)(b+c)(a+c)+abc9=(a+b)(b+c)(a+c)(18−19)−abc9⩾abc9−abc9=0⇒(a+b2)(b+c2)(a+c2)−(a+b+c3)(abc)23⩾0⇒(a+b2)(b+c2)(a+c2)⩾(a+b+c3)(abc)23→(Henceproved)

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