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The AM-GM Inequality

CSCA The AM-GM Inequality study guide organized around the publicly available CSCA syllabus. Practice Mathematics questions on aicsca.com.

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The AM-GM Inequality

In the CSCA exam, the **AM-GM Inequality** (Arithmetic Mean-Geometric Mean) is the nuclear weapon for solving **optimization problems** (Max/Min), especially those involving reciprocal relationships (like $x$ and $\frac{1}{x}$).

1. Core Formula

For any two **positive real numbers** $a, b$ (i.e., $a > 0, b > 0$):

$$\frac{a + b}{2} \ge \sqrt{ab}$$

* **Variation 1 (Minimizing Sum)**: $a + b \ge 2\sqrt{ab}$

* **Variation 2 (Maximizing Product)**: $ab \le (\frac{a + b}{2})^2$

* **Condition for Equality**: The equality holds **if and only if** $a = b$.

2. Geometric Interpretation

The inequality can be proven visually using a semicircle. For a diameter of $a+b$, the radius (Arithmetic Mean) is always greater than or equal to the half-chord height (Geometric Mean).

3. The Three Prerequisites

Memorize these three checks before applying the formula:

1. **Positive**: Terms must be **positive**. If negative, factor out the negative sign.

2. **Constant**: To find a minimum sum, the **product must be constant**; to find a maximum product, the **sum must be constant**.

3. **Equal**: You must verify that the condition $a=b$ is possible within the domain. If $a$ cannot equal $b$, the formula cannot give the exact minimum/maximum.

4. Typical Model: The Hook Function

Functions of the form $f(x) = x + \frac{k}{x} (k>0, x>0)$ are direct applications of AM-GM. The graph looks like a "Nike" checkmark (Hook). Recognizing this shape helps in quick solving.

* Minimum point: When $x = \sqrt{k}$, $y_{min} = 2\sqrt{k}$.

5. Worked Examples

**Ex 1: Basic Reciprocal Model**

Given $x > 0$, find the minimum value of $y = x + \frac{4}{x}$.

**Solution**:

1. **Positive**: $x>0$, check.

2. **Constant**: Product $x \cdot \frac{4}{x} = 4$ (constant).

3. **Apply**: $y \ge 2\sqrt{x \cdot \frac{4}{x}} = 2\sqrt{4} = 4$.

4. **Equal**: Set $x = \frac{4}{x} \Rightarrow x^2=4 \Rightarrow x=2$. Since $x=2$ is in the domain, it works.

**Ans**: Minimum is 4.

**Ex 2: Creating a Constant (Grouping)**

Given $x > 2$, find the minimum value of $y = x + \frac{1}{x-2}$.

**Solution**:

Direct multiplication doesn't work. Rewrite $x$ as $(x-2) + 2$.

$y = (x-2) + \frac{1}{x-2} + 2$

Since $x > 2$, $(x-2) > 0$ (Positive).

Apply AM-GM to the first two terms:

$(x-2) + \frac{1}{x-2} \ge 2\sqrt{(x-2) \cdot \frac{1}{x-2}} = 2$

So $y \ge 2 + 2 = 4$.

Equality check: $(x-2) = \frac{1}{x-2} \Rightarrow x=3$.

**Ex 3: Advanced Splitting (Optional)**

Given $x > 0$, find min of $y = 2x + \frac{8}{x^2}$.

**Note**: Product of terms is $\frac{16}{x}$, not constant. Split $2x$ into $x + x$ to use the **3-term AM-GM**: $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$.

$y = x + x + \frac{8}{x^2} \ge 3\sqrt[3]{x \cdot x \cdot \frac{8}{x^2}} = 3\sqrt[3]{8} = 6$.

Equality holds when $x = x = \frac{8}{x^2} \Rightarrow x=2$.