If x is a positive real number, find the range of values for which 13x 1 3 x + 12 1 2 > 14x

If x is a positive real number, find the range of values for which $\frac{1}{3x}$ + $\frac{1}{2}$ > $\frac{1}{4x}$

Answer Details

$\frac{1}{3x}$ + $\frac{1}{2}$ > $\frac{1}{4x}$
= $\frac{2+3x}{6x}$ > $\frac{1}{4x}$
= 4(2 + 3x) > 6x = 12x2 - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0

= x < 0, x < $\frac{1}{6}$ = x < $\frac{1}{6}$ (solution)

Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > $\frac{1}{6}$
Combining solutions in cases(1) and (2)
= x > 0, x < $\frac{1}{6}$ = 0 < x < $\frac{1}{6}$