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Question Number 15217 by Mr Chheang Chantria last updated on 08/Jun/17

Answered by ajfour last updated on 08/Jun/17

v=ab^2 +a^2 b+ac+bc v =ab(a+b)+c(a+b) =(a+b)(ab+c)≤(a+b)[(((a+b)^2 )/4)+c] 4v ≤(1−c)[(1−c)^2 +4c] 4v≤(1−c)(1+c)^2 [=f(x)] ≤max[f(x)] for 0≤x≤1 ∀ f(x)=(1−x)(1+x)^2 f ′(x)=−(1+x)^2 +2(1+x)(1−x) =−(1+x)(1+x+2x−2) =−3(1+x)(x−(1/3)) f ′(x)=0 at x=−1, (1/3) Since f(−1)=0, f ′(−1)=0, f(0)=1,f ′(0)=1, f ′((1/3))=0, f((1/3))=((32)/(27)) , f(1)=0 , so for 0≤x≤1, max[f(x)]=((32)/(27)) 4v ≤ ((32)/(27)) ⇒ v=a(b^2 +c)+b(a^2 +c)≤(8/(27)) .

v=ab2+a2b+ac+bcv=ab(a+b)+c(a+b)=(a+b)(ab+c)⩽(a+b)[(a+b)24+c]4v⩽(1−c)[(1−c)2+4c]4v⩽(1−c)(1+c)2[=f(x)]⩽max[f(x)]for0⩽x⩽1∀f(x)=(1−x)(1+x)2f′(x)=−(1+x)2+2(1+x)(1−x)=−(1+x)(1+x+2x−2)=−3(1+x)(x−13)f′(x)=0atx=−1,13Sincef(−1)=0,f′(−1)=0,f(0)=1,f′(0)=1,f′(13)=0,f(13)=3227,f(1)=0,sofor0⩽x⩽1,max[f(x)]=32274v⩽3227⇒v=a(b2+c)+b(a2+c)⩽827.

Commented by mrW1 last updated on 08/Jun/17

brilliant method!

brilliantmethod!

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