Chapter 1 · Algebraic Proofs · Concept 5
1-16 Proving Inequalities (4)
A farmer has 200 metres of fence and wants to enclose the biggest field they can. A factory has a fixed amount of metal and wants the roomiest tin. Both questions have the same shape: something is fixed, and you want the best possible. One inequality answers them all.
Guiding question
If two numbers multiply to something fixed, how small can their sum get?
Learning outcome. Prove and use the inequality between the arithmetic mean and the geometric mean.
The mean inequality (AM ≥ GM)
for a ≥ 0 and b ≥ 0: (a + b)/2 ≥ √(ab)
The arithmetic mean is the ordinary average you already know: add them, halve them. The geometric mean is the other kind of middle: multiply them, take the root. The average is never the smaller of the two.
Doubled, it is the form you will use most:
a + b ≥ 2√(ab)
And the sides are equal only when a = b.
The proof is one line, and you have already seen it. a + b − 2√(ab) = (√a − √b)², which is a square, so it is never negative — and it is zero only when √a = √b, that is a = b. That is lesson 1-14 doing the work again.
Keep the product fixed and there is only one thing left to choose: how to split it. Every split gives a different sum — and one split gives the smallest sum there is.
The product a × b is held still at 36, so when you make a bigger, b gets smaller to keep it there. The orange line is the geometric mean — it cannot move. The blue curve is the arithmetic mean.
Push a to either end and the curve climbs away fast. That is the useful part: if you know the product, you know the smallest the sum can ever be.
Nobody will hand you “a + b”. They will hand you something like a + 4/a and expect you to see it. So here is what to look for:
find two pieces whose product has no letters left in it
a + 4/a
a × 4/a = 4
≥ 2√4 = 4
b/a + a/b
(b/a) × (a/b) = 1
≥ 2√1 = 2
x² + 4/x²
x² × 4/x² = 4
≥ 2√4 = 4
3ab + 4/(ab)
3ab × 4/(ab) = 12
≥ 2√12 = 4√3
The letters cancelling is the whole signal. If the product still has an a in it, you have picked the wrong two pieces — try again.
1. If a > 0 and b > 0, prove that b/a + a/b ≥ 2, and say when they are equal.
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Check you are allowed to use the rule.
a > 0 and b > 0, so both pieces are positiveAM ≥ GM only works for non-negative numbers. Say so first, every time.
2. If a, b > 0, prove that (a + b)(1/a + 1/b) ≥ 4
Work with a partner. Before writing anything, both of you say which two pieces you picked and what they multiply to. If you picked differently, check whose product loses the letters.
Prove the following and state when the two sides are equal.
A rectangular field is to be fenced with 200 metres of fencing. Prove that its area can never be more than 2500 m2, and find the shape that reaches it.
Try AM ≥ GM with a = −2 and b = −8. Work out both means. What goes wrong, and why does the rule say a and b must be non-negative?
Work out (a + b)/2 first, then √(ab).
Using the rule without checking the numbers are positive.
The Reflect question shows exactly what that costs. One sentence — "a and b are positive" — protects the whole proof.
Picking two pieces whose product still has letters.
For a + 4/a the pieces are a and 4/a, product 4. Picking a and 4 gives product 4a, which is not a fixed number and proves nothing.
Leaving the answer as 2√12.
2√12 = 2 × 2√3 = 4√3. The book asks for the tidy form, and the mark is for the tidy form.
Getting the equality case from the wrong equation.
The two sides are equal when the two PIECES are equal, not when a = b. For b/(3a) + 12a/b, set b/(3a) = 12a/b, which gives b = 6a — not a = b.
Losing the minus when solving the equality case.
b² = 36a² gives b = 6a or b = −6a. The letters are positive here, so keep only b = 6a.
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