(1/3)x≥(1/3)^1,5 inequation

Given the inequality:
$$\frac{x}{3} \geq \left(\frac{1}{3}\right)^{\frac{3}{2}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x}{3} = \left(\frac{1}{3}\right)^{\frac{3}{2}}$$
Solve:
Given the linear equation:

(1/3)*x = (1/3)^(3/2)

Expand brackets in the left part

1/3x = (1/3)^(3/2)

Expand brackets in the right part

1/3x = 1/3^3/2

Divide both parts of the equation by 1/3

x = sqrt(3)/9 / (1/3)

$$x_{1} = \frac{\sqrt{3}}{3}$$
$$x_{1} = \frac{\sqrt{3}}{3}$$
This roots
$$x_{1} = \frac{\sqrt{3}}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\sqrt{3}}{3}$$
=
$$- \frac{1}{10} + \frac{\sqrt{3}}{3}$$
substitute to the expression
$$\frac{x}{3} \geq \left(\frac{1}{3}\right)^{\frac{3}{2}}$$
$$\frac{- \frac{1}{10} + \frac{\sqrt{3}}{3}}{3} \geq \left(\frac{1}{3}\right)^{\frac{3}{2}}$$

         ___      ___
  1    \/ 3     \/ 3 
- -- + ----- >= -----
  30     9        9  
    

but

         ___     ___
  1    \/ 3    \/ 3 
- -- + ----- < -----
  30     9       9  
   

Then
$$x \leq \frac{\sqrt{3}}{3}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\sqrt{3}}{3}$$

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        /
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       x1