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If-a-3-b-3-a-2-b-2-4-Then-a-4-b-4-2-

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Question Number 205420 by hardmath last updated on 20/Mar/24
If a^3 + b^3 + a^2 + b^2 = 4 Then: a^4 + b^4 ≥ 2
Ifa3+b3+a2+b2=4Then:a4+b4⩾2
Answered by Berbere last updated on 21/Mar/24
2(a^4 +b^4 )+a^2 +b^2 +1+1 =a^4 +a^2 +b^4 +b^2 +a^4 +1+b^4 +1 ≥^(AM−GM) 2∣a^3 ∣+2∣b^3 ∣+2a^2 +2b^2 ≥2(a^3 +b^3 +a^2 +b^2 )=8 2(a^4 +b^4 )+a^2 +b^2 ≥6 a^4 +b^4 ≥^(AM−QM) (((a^2 +b^2 )^2 )/2);X=a^2 +b^2 ≥0 a^4 +b^4 ≥Max(((6−X)/2),(X^2 /2))=(1/2)Max(6−X,X^2 ) X^2 +X−6=(X+3)(X−2) X∈[−3,2] max((X^2 /2),(1/2)(6−X))=(1/2)(6−X)≥((6−2)/2)=2 X≤−3; Max((x^2 /2),((6−x)/2))=(((−3)^2 )/2)≥2 X≥2 Max((X^2 /2),((6−X)/2))=(X^2 /2)≥2 ⇒∀X=a^2 +b^2 ∈[0,∞[ a^4 +b^4 ≥2
2(a4+b4)+a2+b2+1+1=a4+a2+b4+b2+a4+1+b4+1⩾AM−GM2∣a3∣+2∣b3∣+2a2+2b2⩾2(a3+b3+a2+b2)=82(a4+b4)+a2+b2⩾6a4+b4⩾AM−QM(a2+b2)22;X=a2+b2⩾0a4+b4⩾Max(6−X2,X22)=12Max(6−X,X2)X2+X−6=(X+3)(X−2)X∈[−3,2]max(X22,12(6−X))=12(6−X)⩾6−22=2X⩽−3;Max(x22,6−x2)=(−3)22⩾2X⩾2Max(X22,6−X2)=X22⩾2⇒∀X=a2+b2∈[0,∞[a4+b4⩾2
Commented by hardmath last updated on 21/Mar/24
thank you dear professor cool
thankyoudearprofessorcool

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