Inequality question

You can start by sketching a graph of each side and comparing them. That should give you some major insights!

 

Did you make a sketch?

 

Don't we deserve to see your approach?

|x-8| - |2x+1| < 0

This solution is based on the fact that if a continuous function changes its sign, it might happen only in this function zeroes.

1.
Mark zeroes for each |expression| on the number line: x=8 and x=-1/2.
This two points break the whole number line into three intervals:
I. {-inf, -1/2}: -inf < x < 8
II. [-1/2, 8}: -1/2 <= x < 8
III. [8, +inf}: 8<= x < +inf.

Each expression "keeps" its sign unchanged on each interval.
Draw the number line and put the expressions' signs on the intervals
x-8: .............-..........|........-.........|.......+.........
2x+1: ...........-..........|........+.........|.......+.........

Solve inequality on each region separately following the definition
|a| = -a, if a<0
|a| = a, if a>=0

I.
x-8<0 and 2x+1<0 therefore
-(x-8) - (-(2x+1)) < 0
-x + 8 + 2x + 1 < 0
x < -9 - agrees with -inf < x < 8 condition.

II.
x-8<0 and 2x+1>=0 therefore
-(x-8) - (2x+1) < 0
-x + 8 - 2x - 1 < 0
3x > 7
x > 7/3 - considering -1/2 <= x < 8 restriction
the answer on this interval
7/3 < x < 8

III.
x-8>=0 and 2x+1>=0 therefore
x - 8 - (2x+1) < 0
x - 8 - 2x - 1 < 0
x > - 9 - considering x>=8 restriction
the answer on this interval
x >= 8.

Combininig all the answers,
x < -9 or x > 7/3.

Faster way - just like Tide recommended.
Draw two graphs:
f(x) = |x-8| and g(x) = |2x+1|
(do you need help with plotting them?).

|x-8|<|2x+1| is equvalent to
graph f(x) is below graph g(x).
These two graphs intersect at x=-9 and x=7/3.
Graph f(x) is below graph g(x) left of -9 and right of 7/3.
The same answer.

 

Tide just did!

Yes, you did. You did not, however, say whether or not one of the things you tried was graphing the two functions as Tide suggested- that was his question.
In fact, although you say you tried, you have shown us NOTHING of what you tried.
That is, after all, a requirement for homework help on this forum.