AM-GM

a+b/2 ≥ √ab — bounds for products via sums and back.

The idea

Theorem (AM-GM inequality).

For $a, b \geq 0$, $\dfrac{a+b}{2} \;\geq\; \sqrt{ab},$ with equality if and only if $a = b$. More generally, for $n$ nonnegative numbers $a_1, \ldots, a_n$, $\dfrac{a_1 + \cdots + a_n}{n} \;\geq\; \sqrt[n]{a_1 \cdots a_n},$ with equality if and only if all the $a_i$ are equal.

The left side is the arithmetic mean of $a$ and $b$: add them and halve. The right side is the geometric mean: multiply them and take the square root.

AM-GM converts between sums and products: when a product is fixed, it puts a floor under the sum, and the equality condition identifies where that floor is attained. This is what makes it a minimization tool.

Ways to work on it

• Walkthrough. Verify the inequality on a pair, then use it to minimize a sum with a fixed product.
• Proof. See why AM ≥ GM — the semicircle picture.
• Practice. Verify AM-GM on a random pair.
• Hardest. Minimize x + c/x for random c.

Not sure where to start? Take the ten-question placement test.