Question 1209902
Let's analyze the given inequality:

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

Since $x$ is a positive real number, we can multiply both sides of the inequality by $x$ without changing the direction of the inequality:

$$1 \ge 3x - 2x^2$$

Rearrange the terms to get a quadratic inequality:

$$2x^2 - 3x + 1 \ge 0$$

Factor the quadratic expression:

$$(2x - 1)(x - 1) \ge 0$$

To determine when this inequality holds, we need to consider the roots of the quadratic, which are $x = 1/2$ and $x = 1$. We can analyze the sign of the expression $(2x - 1)(x - 1)$ in the intervals defined by these roots.

**Interval Analysis:**

1.  **$x < 1/2$:**
    * $(2x - 1) < 0$ and $(x - 1) < 0$, so $(2x - 1)(x - 1) > 0$.
2.  **$1/2 < x < 1$:**
    * $(2x - 1) > 0$ and $(x - 1) < 0$, so $(2x - 1)(x - 1) < 0$.
3.  **$x > 1$:**
    * $(2x - 1) > 0$ and $(x - 1) > 0$, so $(2x - 1)(x - 1) > 0$.

Also, the expression is equal to 0 when $x=1/2$ and $x=1$.

Therefore, the inequality $(2x - 1)(x - 1) \ge 0$ holds when $x \le 1/2$ or $x \ge 1$.

Since we are given that $x$ is a positive real number, we have:

* $0 < x \le 1/2$ or $x \ge 1$

**Equality Condition:**

Equality occurs when $(2x - 1)(x - 1) = 0$, which means either $2x - 1 = 0$ or $x - 1 = 0$.

* $2x - 1 = 0 \Rightarrow x = 1/2$
* $x - 1 = 0 \Rightarrow x = 1$

**Conclusion:**

The inequality $\frac{1}{x} \ge 3 - 2x$ holds for all positive real numbers $x$ such that $0 < x \le 1/2$ or $x \ge 1$.

Equality occurs when $x = 1/2$ or $x = 1$.