Algebra

The inequality 1x≥3−2x\frac{1}{x} \ge 3 - 2xx1​≥3−2x holds for x∈(0,12]∪[1,∞) x \in (0, \frac{1}{2}] \cup [1, \infty) x∈(0,21​]∪[1,∞), and equality occurs at x=12 x = \frac{1}{2} x=21​ and x=1 x = 1 x=1. It does not hold for all x>0 x > 0 x>0 (e.g., fails at x=0.75 x = 0.75 x=0.75), so the statement is true only in these intervals.

$\frac{1}{x} \ge 3 - 2x$

1  ≥ 3x - 2x^2

2x^2 -3x + 1  ≥  0       set as an equality  to find test intervals

2x^2 - 3x + 1   =  0

x =   [ 3 + sqrt [ 9 - 8]] / 4  =  1     or    x =  [ 3 - sqrt [ 9 -8 ] ] / 4] =  1/2

Solution

0 < x  ≤ 1/2   ∪    x  ≥ 1