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Question Number 151596 by mathdanisur last updated on 22/Aug/21

Answered by dumitrel last updated on 22/Aug/21

((√a)+(√b)+(√c))^2 ≤^(cbs) 3(a+b+c)⇒(√a)+(√b)+(√c)≤(√(3(a+b+c))) ((√(ab))+(√(bc))+(√(ac)))^2 ≤^(cbs) 3(ab+bc+ac)⇒(√(ab))+(√(bc))+(√(ac))≤(√(3(ab+bc+ac)) ⇒((√a)+(√b)+(√c))((√(ab))+(√(bc))+(√(ac))≤(√(3(a+b+b)∙3(ab+bc+ac)))= 3(√((a+b+c)(ab+bc+ac)))

(a+b+c)2⩽cbs3(a+b+c)⇒a+b+c⩽3(a+b+c)(ab+bc+ac)2⩽cbs3(ab+bc+ac)⇒ab+bc+ac⩽3(ab+bc+ac⇒(a+b+c)(ab+bc+ac⩽3(a+b+b)⋅3(ab+bc+ac)=3(a+b+c)(ab+bc+ac)

Commented by mathdanisur last updated on 22/Aug/21

Thank you dear Ser, nice

ThankyoudearSer,nice

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