To solve the inequality 3x + 1/x + 2 ≤ 1, we can start by getting rid of the fraction by multiplying both sides of the inequality by x:

x(3x + 1/x + 2) ≤ x(1)

This simplifies to:

3x^2 + 1 + 2x ≤ x

Now, let's rearrange the terms to get a quadratic inequality:

3x^2 + 2x - x + 1 ≤ 0

Combine like terms:

3x^2 + x + 1 ≤ 0

To solve this quadratic inequality, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 3, b = 1, and c = 1. Plugging these values into the quadratic formula, we get:

x = (-1 ± √(1 - 4(3)(1))) / (2(3))

Simplifying further:

x = (-1 ± √(1 - 12)) / 6

x = (-1 ± √(-11)) / 6

Since the discriminant (b^2 - 4ac) is negative, the quadratic equation has no real solutions. Therefore, there are no values of x that satisfy the inequality 3x + 1/x + 2 ≤ 1.