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Question Number 144823 by loveineq last updated on 29/Jun/21

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

Leta,b>0anda+b+1=3ab.Provethata+1b+1+b+1a+1⩽a+b

Answered by ArielVyny last updated on 30/Jun/21

we know that a+b+1=3ab we suppose ab≥1 (a+1)=3ab−b (a+1)^2 =b^2 (3a−1)^2 =b^2 (9a^2 +1−9a) a+1≤9ab (1) →b+1≤9ab (2) (((1))/((2)))+(((2))/((1)))→((a+1)/(b+1))+((b+1)/(a+1))≤2 then 3≤3ab→2≤3ab−1 we have ((a+1)/(b+1))+((b+1)/(a+1))≤2≤3ab−1=a+b ((a+1)/(b+1))+((b+1)/(a+1))≤2≤a+b finally ((a+1)/(b+1))+((b+1)/(a+1))≤a+b (ab≥1)

weknowthata+b+1=3abwesupposeab⩾1(a+1)=3ab−b(a+1)2=b2(3a−1)2=b2(9a2+1−9a)a+1⩽9ab(1)→b+1⩽9ab(2)(1)(2)+(2)(1)→a+1b+1+b+1a+1⩽2then3⩽3ab→2⩽3ab−1wehavea+1b+1+b+1a+1⩽2⩽3ab−1=a+ba+1b+1+b+1a+1⩽2⩽a+bfinallya+1b+1+b+1a+1⩽a+b(ab⩾1)

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