Question 981360
If you mean the way you wrote it, then


{{{x+4/x-2<2/x+1}}}


{{{x+4/x-2-2/x-1<0}}}


{{{(x^2+4-2x-2-x)/x<0}}}


{{{(x^2-3x+2)/x<0}}}


{{{((x-1)(x-2))/x<0}}}
The two zeros and the undefined x value give three critical values which form the x-number line into four intervals:
The critical values are  0, 1, 2.
The intervals on the x-axis are  (-infinity,0), (0,1), (1,2), (2, infinity).


Test any value in each of the intervals and find if it satisfies or not, the inequality.



{{{x+4/x-2<2/x+1}}}
Trying x=-1,
{{{-1+4/(-1)-2<2/(-1)+1}}}
{{{-1-4-2<-2+1}}}
{{{-7<-1}}}
TRUE for (-infinity,0).


{{{x+4/x-2<2/x+1}}}
Try {{{1/2}}},
{{{1/2+8-2<4+1}}}
{{{6&1/2<5}}}
FALSE for (0,1).


Try  {{{3/2}}},
{{{x+4/x-2<2/x+1}}}
{{{3/2+2*4/3-2<3+1}}}
{{{3/2+8/3-2<4}}}
{{{-1/2+8/3<4}}}
{{{2&2/3-1/2<4}}}
TRUE for (1,2).


Try 4,
{{{x+4/x-2<2/x+1}}}
{{{4+1-2<1/2+1}}}
{{{3<3/2}}}
FALSE for (2, infinity)


This is not yet completely solved.  The value for x at 0 cannot be used or included, but the roots of the numerator in the single expression mean that x at 1 and x at 2 SHOULD still be checked, just to be sure of any more thorough solution.


You could try that...