1. The AM-GM inequality states that for any list of non-negative real numbers $a_1, a_2, \dots, a_n$, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM): $$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdots a_n}$$ with equality if and only if all $a_i$ are equal. 2. To prove this, we start with the simplest case $n = 2$: $$\frac{a_1 + a_2}{2} \geq \sqrt{a_1 a_2}$$ 3. Use the fact that the square of any real number is non-negative: $$ (\sqrt{a_1} - \sqrt{a_2})^2 \geq 0$$ which expands to $$a_1 - 2\sqrt{a_1 a_2} + a_2 \geq 0$$ 4. Rearranging gives $$a_1 + a_2 \geq 2 \sqrt{a_1 a_2}$$ Dividing both sides by 2, we get the AM-GM inequality for two numbers. 5. For general $n$, the proof uses induction or Jensen's inequality, but the key idea is similar: the inequality holds and equality occurs if and only if all $a_i$ are equal. Therefore, the AM-GM equality case holds only when all numbers are identical, which completes the proof.