Two ways to average two positive numbers — and why the gap between them unlocks "largest" and "smallest."
There is more than one way to average two numbers, and the two ways quietly disagree. The everyday average of 4 and 16 is their sum cut in half: (4 + 16) ÷ 2 = 10. But there is a second, sneakier "middle" — the geometric mean, the number you'd multiply by itself to land between them: √(4 · 16) = √64 = 8. Ten and eight. The plain average came out larger. That is not a fluke. For any two positive numbers, the ordinary average is never smaller than the geometric mean — and the size of the gap between them turns out to be a master key for finding the largest product, or the smallest sum, that a quantity can possibly reach.
By the end of this lesson you'll be able to state the basic inequality (a + b)/2 ≥ √(ab) for positive numbers, see why it must be true (it is just a squared thing being ≥ 0), recognize it in a semicircle, name the one moment the two sides are equal, and use all of that to answer "how big can it get?" and "how small can it get?" questions. As always, color carries meaning: blue is the structure and the variable, amber is the bigger quantity (the arithmetic mean), green is the smaller one (the geometric mean) and the moment of equality, and red flags a trap — a place people misuse the rule.
Take two positive numbers, a and b. The arithmetic mean is the one you already know — add them and halve: (a + b)/2. The geometric mean is what you get by multiplying and taking the square root: √(ab). Both are honest "middles." Both sit between a and b. But they are not the same number, and one of them always wins.
The claim — the basic inequality, often called AM–GM — is this:
Why must this be true? Not by magic — by a square. A square of a real number is never negative. Start from the most innocent fact in algebra, (√a − √b)² ≥ 0, and just expand it:
The middle line is the heart of it: (√a − √b)² equals a − 2√(ab) + b, and since that whole thing is ≥ 0, the 2√(ab) can be carried to the other side. One squared quantity does all the work.
For positive a and b: (a + b)/2 ≥ √(ab), the same as a + b ≥ 2√(ab). It comes straight from (√a − √b)² ≥ 0. The arithmetic mean is the champion; the geometric mean can tie it but never beat it.
There is a close cousin that needs no square roots and works for any real numbers, positive or not. Start from (a − b)² ≥ 0 and expand: a² − 2ab + b² ≥ 0, so
a² + b² ≥ 2ab.
You met this one already, in a smaller costume. Back in 12.1 we compared a² + 1 with 2a and found a² + 1 − 2a = (a − 1)² ≥ 0, so a² + 1 ≥ 2a for every number, with equality only at a = 1. That was just a² + b² ≥ 2ab with b = 1. The basic inequality is the same instinct grown up.
Arithmetic mean (4 + 16)/2 = 10. Geometric mean √(4 · 16) = √64 = 8. And indeed 10 > 8. The gap is 2. (Try a = b = 9: both means come out 9 — the gap shuts.)
Algebra tells you the inequality is true. A picture tells you why you should have believed it all along. Draw a half-circle sitting on a flat line, and split the diameter into two pieces of length a and b. The diameter is then a + b, so the radius is exactly (a + b)/2 — the arithmetic mean, hiding in plain sight.
Now stand a vertical line straight up from the split point until it hits the curve. That half-chord has a famous length: in a semicircle, the height raised at the split point is the geometric mean of the two pieces — exactly √(ab). (It's the same fact as: the altitude to the hypotenuse of a right triangle is the geometric mean of the two segments it makes.)
Here is the punch line. Every point on the curve is exactly one radius from the center, so the slanted segment from the center out to the top of our upright is a full radius — length (a + b)/2. The green upright is just the vertical leg of the little right triangle whose hypotenuse is that radius. And a leg of a right triangle is never longer than its hypotenuse. So the geometric mean √(ab) can never exceed the radius (a + b)/2 — which is the basic inequality, drawn instead of derived.
Radius (a + b)/2 = the slant (the hypotenuse). Half-chord √(ab) = the upright (a leg). Leg ≤ hypotenuse, so √(ab) ≤ (a + b)/2. Slide the split toward the center and the upright grows until — only at the center — it stands up as a full radius and the two are equal.
So far we've said the amber side wins or ties. The tie is where all the power lives, so pin it down exactly. Look back at the proof: the only slack in the whole argument was the step (√a − √b)² ≥ 0. A square equals zero only when the thing inside is zero. So (√a − √b)² = 0 means √a = √b, which means
the two means are equal if and only if a = b.
That single "equal exactly when a = b" is what lets the inequality answer optimization questions. An inequality alone only says "the value is at least this much" or "at most this much." To claim a true minimum or maximum, you must show that the bound is actually reached — and it is reached precisely at a = b. If a = b is impossible in your problem, the bound is a wall you never quite touch, not a maximum or minimum you attain.
That gives us a three-word checklist before ever trusting the basic inequality to deliver an extreme. Chant it: positive, fixed, equal.
Suppose x ≥ 5 and you write x + 4/x ≥ 2√4 = 4. True as an inequality — but the "4" needs x = 2, which is forbidden here. So 4 is never reached; it is not the minimum on x ≥ 5. Condition ③ failed. Ticking only two of three is the classic mistake.
Now the payoff. The basic inequality is a two-sided machine, and which side you read depends on what's held constant.
From a + b ≥ 2√(ab), square both sides (both are positive): (a + b)² ≥ 4ab, so ab ≤ ((a + b)/2)². If the sum is locked, the product ab can only climb so high — and it hits the ceiling exactly when a = b.
With a + b = 10 fixed, ab ≤ (10/2)² = 5² = 25. The biggest possible product is 25, reached when a = b = 5. Check the three: terms positive ✓, sum fixed at 10 ✓, and a = b = 5 is allowed ✓. So 25 is a genuine maximum.
Read the same machine the other way. From a + b ≥ 2√(ab), if the product is locked at ab = P, then a + b ≥ 2√P. The sum can only sink so low — bottoming out when a = b = √P.
With ab = 16 fixed, a + b ≥ 2√16 = 2 · 4 = 8. The smallest possible sum is 8, reached when a = b = 4 (and indeed 4 · 4 = 16 ✓). All three conditions hold, so 8 is a true minimum.
The most common shape you'll meet is a number plus a constant over that number. For x > 0, the two terms x and k/x have a fixed product: x · (k/x) = k, a constant! So their sum is bounded below:
| expression | minimum (2√k) | reached at x = √k |
|---|---|---|
| x + 1/x | 2 | x = 1 |
| x + 4/x | 4 | x = 2 |
| x + 9/x | 6 | x = 3 |
The flagship case x + 1/x ≥ 2 (least value 2 at x = 1) is worth memorizing. And note x + 4/x ≥ 2√4 = 4, reached at x = 2 (since there x = 4/x). Don't guess that the minimum is at x = 1 for x + 4/x — that gives 5, not the minimum. The minimum lives where the two terms balance.
The minimum of x + k/x is not always at x = 1. It's at x = √k, where the two terms are equal. For x + 9/x the balance point is x = 3, giving the minimum 6 — plugging in x = 1 would wrongly suggest 10.
For positive a and b, the ordinary average never falls below the geometric mean: (a + b)/2 ≥ √(ab), the same as a + b ≥ 2√(ab). It's true because (√a − √b)² ≥ 0 — a square is never negative — and its no-roots cousin is a² + b² ≥ 2ab from (a − b)² ≥ 0. A semicircle draws it for you: the radius is the arithmetic mean, the half-chord is the geometric mean, and a leg is never longer than the hypotenuse. The two sides are equal only when a = b, which is exactly the moment a bound becomes a true extreme. So to optimize, chant positive, fixed, equal: a fixed sum gives the biggest product (a + b = 10 ⇒ ab ≤ 25 at 5, 5); a fixed product gives the smallest sum (ab = 16 ⇒ a + b ≥ 8 at 4, 4); and the flagship x + k/x ≥ 2√k bottoms out at x = √k (so 2, 4, 6 at x = 1, 2, 3 for k = 1, 4, 9).
This rule was built for word problems. In 12.7 you'll fence the largest pen with a fixed length of wire — the same "fixed sum → biggest product" idea, now made of fence — and translate "at least / at most" into inequalities you can solve. The basic inequality is your sharpest tool for "as big as possible" and "as small as possible."
Assume every letter stands for a positive number unless stated otherwise. Use a real minus sign and keep your reasoning honest about the three conditions.
Six questions to lock it in. Tap the answer you think is right.
This lesson is enrichment beyond the core Common Core list — the AM–GM inequality is standard in honors precalculus and competition tracks rather than in the baseline standards. It is best framed against CCSS.HSA-SSE.A (interpret and use the structure of an expression — here, recognizing that x · (k/x) is a constant), CCSS.HSF-IF.C (locate maximum and minimum values of a function), and the Standards for Mathematical Practice, especially MP7 (look for structure) and MP3 (construct a valid argument — the proof is a single squared term). It pairs naturally with the quadratic-vertex view of max/min from Stage 11 and with the modeling problems in 12.7; it builds directly on the difference-method idea from 12.1 and the legal moves from 12.2.
The #1 misconception: using the inequality without checking its conditions — applying it to negative or zero terms, applying it when the sum or product isn't actually constant, or declaring a "minimum/maximum" at a point where a = b can't be reached (so the bound is never attained). The antidote is the chant "positive, fixed, equal": confirm the terms are positive, confirm the relevant sum or product is genuinely held constant, and confirm the equality case a = b truly lives inside the allowed domain. If the third box can't be ticked, the bound is a wall, not an extreme — find the real extreme another way (often at the domain's edge).
Nearby in Stage 12: 12.1 · 12.2 · 12.3 · 12.4 · 12.5 · 12.6 · 12.7