The whole lesson in one picture: the average of two lengths (amber radius, the slant from the center) is never smaller than their geometric mean (green half-chord, the upright leg) — a leg never outruns its hypotenuse. The two match only when the split lands dead center — when a = b.
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.
12.6.1 Two averages, and the tug-of-war between them
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:
The arithmetic mean (amber) is never below the geometric mean (green). Doubling both sides gives the form a + b ≥ 2√(ab), which is the version we'll lean on for optimizing.
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:
Three honest lines. The whole inequality is just "a square is never negative," rearranged. Halving the last line gives (a + b)/2 ≥ √(ab).
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.
Key idea
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.
Example · a = 4, b = 16
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.)
🎮 Try it AM vs GM — two bars
Set two positive numbers. The amber bar is their ordinary average; the green bar is their geometric mean. Watch the amber bar stay ahead — and see the gap snap shut the instant the two numbers are equal.
a4
b16
12.6.2 The geometric picture: a semicircle never lies
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.)
The green upright is a vertical leg; the amber radius is the slant from the center to the same point on the curve. A leg of a right triangle is never longer than the hypotenuse, so √(ab) ≤ (a + b)/2. They tie only when the split is at the center.
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.
Read it off the picture
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.
🎮 Try it Drag the split in a semicircle
Slide the split point along the diameter. The amber radius never changes length; the green half-chord grows toward it and reaches it only when the split is dead center.
split
12.6.3 The one moment of equality — "positive, fixed, 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 ifa = 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.
Three boxes you must tick before the bound counts as a real maximum or minimum.
Watch out — the equality trap
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.
🎮 Try it Does the bound actually count?
Each card is an attempt to use the basic inequality. Decide whether all three of positive · fixed · equal are met. Tap your verdict; the widget checks it and says which condition (if any) breaks.
12.6.4 Finding extremes: biggest product, smallest sum
Now the payoff. The basic inequality is a two-sided machine, and which side you read depends on what's held constant.
Fixed sum → biggest product
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.
Pinned · a + b = 10
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.
Fixed product → smallest sum
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.
Pinned · ab = 16
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 flagship: x + k/x
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:
The two terms x and k/x multiply to the constant k, so their sum is smallest when they're equal — that is, when x = √k.
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.
Watch out
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.
🎮 Try it Minimize x + k/x
Pick k, then slide x > 0. The curve is x + k/x; the dashed line is the floor 2√k. Find where the moving dot touches the floor — that's the minimum, at x = √k.
k
x
★ The big ideas, in one breath
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).
What's next
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."
✎ Exercises 12.6
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.
Compute both means of a = 2 and b = 8. Which is larger, and by how much?
Show answer
Arithmetic mean (2 + 8)/2 = 5. Geometric mean √(2 · 8) = √16 = 4. The arithmetic mean is larger, by 1. (As always, AM ≥ GM.)
Fill the blank with one of < = >: for any real numbers, a² + b² ___ 2ab. Justify it in one line.
Show answer
≥. Because a² + b² − 2ab = (a − b)² ≥ 0. Equality only when a = b.
Two positive numbers add to 10. What is the largest their product can be, and at what values?
Show answer
Fixed sum → biggest product. ab ≤ ((a + b)/2)² = (10/2)² = 25, reached when a = b = 5. Check: 5 · 5 = 25 ✓; conditions positive·fixed·equal all hold.
Two positive numbers multiply to 16. What is the smallest their sum can be, and where?
Show answer
Fixed product → smallest sum. a + b ≥ 2√(ab) = 2√16 = 8, reached when a = b = 4 (4 · 4 = 16 ✓). So the minimum sum is 8.
Find the least value of x + 1/x for x > 0, and the x where it happens.
Show answer
The terms x and 1/x have product x · (1/x) = 1 (fixed). So x + 1/x ≥ 2√1 = 2, reached when x = 1/x, i.e. x = 1. Least value 2.
Find the least value of x + 4/x for x > 0. (Hint: don't assume it's at x = 1.)
Show answer
Product x · (4/x) = 4 (fixed). So x + 4/x ≥ 2√4 = 4, reached when x = 4/x ⇒ x² = 4 ⇒ x = 2. Least value 4 at x = 2. (At x = 1 you'd get 5 — bigger, not the minimum.)
Find the least value of x + 9/x for x > 0, with the location.
Show answer
Product x · (9/x) = 9. So x + 9/x ≥ 2√9 = 6, reached when x = 9/x ⇒ x² = 9 ⇒ x = 3. Least value 6 at x = 3.
A student writes: "For all real x, x + 1/x ≥ 2." What's wrong, and how would you fix the statement?
Show answer
The basic inequality needs positive terms. If x < 0 the claim fails — e.g. x = −1 gives x + 1/x = −2, which is not ≥ 2. The correct statement is "for x > 0, x + 1/x ≥ 2." (For x < 0 one finds instead x + 1/x ≤ −2.)
On a semicircle, the diameter is split into pieces of length 9 and 4. How long is the upright half-chord at the split, and how long is the radius? Which is bigger?
Show answer
Half-chord = √(9 · 4) = √36 = 6 (geometric mean). Radius = (9 + 4)/2 = 6.5 (arithmetic mean). The radius is bigger, as it must be — the split isn't at the center, so the upright is shorter than a full radius.
Try to use the basic inequality on x + 4/x when x ≥ 5. Is the minimum still 4? Explain using "positive, fixed, equal."
Show answer
No. Positive ✓ and fixed product 4 ✓, but equal fails: equality needs x = 2, which is outside x ≥ 5. So 4 is never reached on this domain — it isn't the minimum here. (On x ≥ 5 the expression is increasing, so its least value is at the left edge, x = 5: 5 + 4/5 = 5.8.)
🎯 Quick check
Six questions to lock it in. Tap the answer you think is right.
§ For teachers and parents
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).