正数不等式证明求助：求证$x + \frac{1}{4x} \geq 1$（x>0）

Hey Rick, let's work through this inequality proof together! Since we know (x) is a positive real number ((x > 0)), there are two straightforward, easy-to-follow methods to show (x + \frac{1}{4x} \geq 1).

方法1：算术-几何均值不等式（AM-GM） #

For any pair of positive real numbers (a) and (b), the arithmetic mean is always greater than or equal to the geometric mean:
$$\frac{a + b}{2} \geq \sqrt{ab}$$

Let’s set (a = x) and (b = \frac{1}{4x}) (both are positive because (x > 0)). Plug these into the AM-GM formula:
$$\frac{x + \frac{1}{4x}}{2} \geq \sqrt{x \cdot \frac{1}{4x}}$$

First simplify the right-hand side:
$$\sqrt{x \cdot \frac{1}{4x}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$

Multiply both sides of the inequality by 2:
$$x + \frac{1}{4x} \geq 1$$

Equality holds when (a = b), so (x = \frac{1}{4x}). Solving this gives (4x^2 = 1), and since (x > 0), (x = \frac{1}{2}).

方法2：配方法 #

We can rearrange the inequality to form a perfect square (which is always non-negative). Start with the original inequality:
$$x + \frac{1}{4x} \geq 1$$

Multiply both sides by (4x) (since (x > 0), the inequality direction stays the same):
$$4x^2 + 1 \geq 4x$$

Move all terms to one side:
$$4x^2 - 4x + 1 \geq 0$$

Notice the left-hand side is a perfect square:
$$(2x - 1)^2 \geq 0$$

Since the square of any real number is non-negative, this inequality is always true for all real (x) (including positive (x)). Equality holds when (2x - 1 = 0), so (x = \frac{1}{2}).

Either method works perfectly to prove the inequality! 😊

内容的提问来源于stack exchange，提问作者Rick