Simple absolute value inequalityProve: for all x ≠ 0 , | x + 1 x | &gt; 2 and | x + 1 x | = 2 ⟺ x = ± 1case 1: x &gt; 0 x + 1 x &gt; 2 x 2 − 2 x + 1 &gt; 0 x 1 , 2 = 2 + 4 − 4 2 = 1Therefore x ≠ 1Case 2: x &lt; 0In the same manner we get x 2 + 2 x + 1 &gt; 0and x ≠ − 1Now 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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Simple absolute value inequalityProve: for all   x  ≠  0  ,      |    x    +          1      x        |    &amp;gt;  2 and       |    x    +          1      x        |    =  2    ⟺    x  =  ±  1case 1:   x  &amp;gt;  0  x  +      1    x    &amp;gt;  2      x    2    −  2  x  +  1  &amp;gt;  0      x          1      ,      2        =            2      +              4        −        4              2    =  1Therefore   x  ≠  1Case 2:   x  &amp;lt;  0In the same manner we get      x    2    +  2  x  +  1  &amp;gt;  0and   x  ≠  −  1Now we need that both cases will occur simultaneously (?) so we get   x  =≠  ±  1 for an equality and   x  ≠  0 for inequality

Jayleen Copeland · Accepted Answer

It&#039;s just AM-GM:                    x        2            +      1        2    ≥      |    x      |    ,which for   x  ≠  0 gives which you wish:                    x        2            +      1                      |            x              |              ≥  2or      |    x    +          1      x        |    ≥  2.

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