Solve the inequation: `|1-(|x|/(1+|x|))|>=1/2`

Explanation

The given inequation is equivalent to the collectionn of systems
`{(|1-x/(1+x)|ge1/2, "if"xge0),(|1+x/(1-x)|ge1/2, "if"xlt0):}implies{(`/(|1+x|)ge1/2, "if"xge0),(1/(|1-x|)ge1/2, "if"xlt0):}`
`implies{(1/(1+x)ge1/2, "if"xge0),(1/(1-x)ge1/2, "if"xlt0):}implies{((1-x)/(1+x)ge0, "if"xge0),((1+x)/(1-x)ge0,"if"xlt0):}`
`implies [((x-1)/(x+1)le0, "if" xge0),((x+1)/(x-1)le0, "if"xlt0):}`
For `(x-1)/(x+1)le0,` if `xge0`

`:.-1lexlt0`.............ii
Hence from Eqs i and ii the solution of the given equation is `x epsilon [-1,1]`
Aliter `|1-(|x|)/(1+|x|)ge1/2implies|1/(1+|x|)ge1/2`
`implies1/(1+|x|)ge1/2implies1+|x|le2` or `|x|le1`
`:.-1lexle1` or `ximplies[-1,1]`