SOLUTION: Let x be a positive real number. Show that \frac{1}{x} \ge 3

SOLUTION: Let x be a positive real number. Show that \frac{1}{x} \ge 3 - 2x Describe when we have equality.

Question 1209902: Let x be a positive real number. Show that
\frac{1}{x} \ge 3 - 2x
Describe when we have equality.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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 $x = 1$.

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