Simple absolute value inequality Prove: for all x!=0, ∣∣x+1/x∣∣>2 and ∣∣x+1/x∣∣=2<=>x=+-1 case 1: x>0 x+1/x>2 x^2−2x+1>0 x_(1,2)=((2+sqrt(4−4))/2=1 Therefore x!=1 Case 2: x<0 In the same manner we get x^2+2x+1>0 and x!=−1 Now we need that both cases will occur simultaneously (?) so we get x!=+-1 for an equality and x!=0 for inequality

hazbijav6 2022-09-05 Answered
Simple absolute value inequality
Prove: for all x 0 , | x + 1 x | > 2 and | x + 1 x | = 2 x = ± 1
case 1: x > 0
x + 1 x > 2
x 2 2 x + 1 > 0
x 1 , 2 = 2 + 4 4 2 = 1
Therefore x 1
Case 2: x < 0
In the same manner we get
x 2 + 2 x + 1 > 0
and x 1
Now we need that both cases will occur simultaneously (?) so we get x =≠ ± 1 for an equality and x 0 for inequality
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Answers (1)

Jayleen Copeland
Answered 2022-09-06 Author has 4 answers
It's just AM-GM:
x 2 + 1 2 | x | ,
which for x 0 gives which you wish:
x 2 + 1 | x | 2
or
| x + 1 x | 2.
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