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Question Number 144537 by loveineq last updated on 26/Jun/21

Let a,b>0 and a+b = 2. Prove that (1) ((a^3 +b^3 )/2)−2(1−ab) ≥ 1 (2) ((a^2 +b^2 )/2)−2(1−ab) ≤ 1

Leta,b>0anda+b=2.Provethat (1)a3+b32−2(1−ab)⩾1 (2)a2+b22−2(1−ab)⩽1

Answered by mnjuly1970 last updated on 26/Jun/21

(1) : (((a+b)(a^( 2) +b^( 2) −ab))/2)−2+2ab =a^( 2) +b^2 −2+ab=(a+b)^2 −ab−2 =2−ab ≥_(inequality) ^(am −gm) 2−((a+b)/2) =1 ...

(1):(a+b)(a2+b2−ab)2−2+2ab =a2+b2−2+ab=(a+b)2−ab−2 =2−ab⩾am−gminequality2−a+b2=1...

Commented byloveineq last updated on 26/Jun/21

thanks

thanks

Answered by Olaf_Thorendsen last updated on 26/Jun/21

(1) : x = ((a^3 +b^3 )/2)−2(1−ab) x = (((a+b)^3 −3ab(a+b))/2)−2(1−ab) x = ((8−6ab)/2)−2(1−ab) x = 2−ab = 2−a(2−a) x = a^2 −2a+2 = (a−1)^2 +1 ≥ 1 (2) y = ((a^2 +b^2 )/2)−2(1−ab) y = (((a+b)−2ab)/2)−2(1−ab) y = ((2−2ab)/2)−2(1−ab) y = ab−1 = a(2−a)−1 y = −a^2 +2a−1 = −(a−1)^2 ≤ 0 ≤ 1

(1):x=a3+b32−2(1−ab) x=(a+b)3−3ab(a+b)2−2(1−ab) x=8−6ab2−2(1−ab) x=2−ab=2−a(2−a) x=a2−2a+2=(a−1)2+1⩾1 (2)y=a2+b22−2(1−ab) y=(a+b)−2ab2−2(1−ab) y=2−2ab2−2(1−ab) y=ab−1=a(2−a)−1 y=−a2+2a−1=−(a−1)2⩽0⩽1

Commented byloveineq last updated on 26/Jun/21

thanks

thanks

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