Menu Close

Let-a-b-0-and-a-1-b-1-a-b-2-Prove-that-a-b-a-1-3-b-1-3-a-1-2-b-1-2-1-2-a-1-3-b-1-3-




Question Number 145314 by loveineq last updated on 04/Jul/21
Let a,b ≥ 0 and (a+1)(b+1) = (a+b)^2  . Prove that         (a+b)(√((a+1)^3 +(b+1)^3 )) ≤ (a+1)^2 +(b+1)^2  ≤ (1/2)[(a+1)^3 +(b+1)^3 ]
Leta,b0and(a+1)(b+1)=(a+b)2.Provethat(a+b)(a+1)3+(b+1)3(a+1)2+(b+1)212[(a+1)3+(b+1)3]
Commented by justtry last updated on 05/Jul/21
(a+b)(b+1)=(a+b)^2   ⇔ab+a+b+1=a^2 +2b+b^2   ⇔1+(a+b)=a^2 +2b+b^2   b=0⇒(1+a)=a^2   a=0⇒(1+b)=b^2   & (1+a)^3 ≥(1+a)^2 =(a^2 )^2         (1+b)^3 ≥(1+b)^2 =(b^2 )^2   (1/2)[(1+a)^3 +(1+b)^3 ]≥(1/2)[(a^2 )^2 +(b^2 )^2 ]≥(((a^2 +b^2 )^2 )/4)  ⇔(1/2)[(1+a)^3 +(1+b)^3 ]≥((a^2 )^2 +(b^2 )^2 )≥(((a^2 +b^2 )^2 )/2)  ⇔ (1/2)[(1+a)^3 +(1+b)^3 ]≥(1/2)(a^4 +b^4 +2a^2 b^2 )=(1/2)[(1+a)^2 +(1+b)^2 +2(a+b)^2 ]  note  (1/2)[(1+a)^2 +(1+b)^2 +2(a+b)^2 ]≥(1+a)^2 +(1+b)^2  →a=b=1 or a≠b≠1, a,b≥0  so  (1/2)[(1+a)^3 +(1+b)^3 ]≥(1+a)^2 +(1+b))^2   ⇔ (1+a)^2 +(1+b)^2 ≤(1/2)[(1+a)^3 +(1+b)^3 ]  .........
(a+b)(b+1)=(a+b)2ab+a+b+1=a2+2b+b21+(a+b)=a2+2b+b2b=0(1+a)=a2a=0(1+b)=b2&(1+a)3(1+a)2=(a2)2(1+b)3(1+b)2=(b2)212[(1+a)3+(1+b)3]12[(a2)2+(b2)2](a2+b2)2412[(1+a)3+(1+b)3]((a2)2+(b2)2)(a2+b2)2212[(1+a)3+(1+b)3]12(a4+b4+2a2b2)=12[(1+a)2+(1+b)2+2(a+b)2]note12[(1+a)2+(1+b)2+2(a+b)2](1+a)2+(1+b)2a=b=1orab1,a,b0so12[(1+a)3+(1+b)3](1+a)2+(1+b))2(1+a)2+(1+b)212[(1+a)3+(1+b)3]
Commented by loveineq last updated on 05/Jul/21
  why 1+a+b=a^2 +2b+b^2 ?
why1+a+b=a2+2b+b2?

Leave a Reply

Your email address will not be published. Required fields are marked *