Chapter 1 · Algebraic Proofs · Concept 5
1-13 Proving Inequalities (1)
A bio-engineer is designing a capsule to carry a drug to a tumour. They want the largest volume they can get, but there is a hard limit on how much of the expensive outer polymer they may use. An equation cannot answer that. An inequality can — it tells you the best that is possible.
Guiding question
How do you prove that one expression is always bigger than another?
Learning outcome. Prove that algebraic inequalities are correct in various math situations (conditional).
The whole method
To prove LHS > RHS, work out LHS − RHS and show it is positive.
If you are given a condition like a > b, rewrite it as a − b > 0 so you can use it.
Why the subtraction? Because “bigger than” is hard to work with, and “positive” is easy. A positive number is one you can recognise by factorising — and every question on these pages factorises into brackets you already know are positive.
The difference always looks messy at first and always tidies into two brackets. Then the condition tells you each bracket is positive, and you are finished.
- LHS − RHS = 2(ax + by) − (a + b)(x + y)
- …
1. If a > b and x > y, prove that 2(ax + by) > (a + b)(x + y)
- 1
Start by writing LHS − RHS.
2(ax + by) − (a + b)(x + y)Always this first line. Do not move anything across the inequality sign — the rule from lesson 1-10 still applies.
Work with a partner. Do these together, then talk about the answers. Write LHS − RHS first, before anything else.
Prove the following inequalities.
If a > b > c > 0 and x > y > z > 0, prove that ax + by + cz > ay + bz + cx.
Never move a term across the sign.
Exactly as in lesson 1-10. Rearranging the inequality assumes it is already true, which is the thing you were asked to prove.
Turn every condition into "something > 0".
a > b is not directly usable. a − b > 0 is, because that is the shape the factors come out in.
Group the terms, do not stare at them.
ax − ay − bx + by looks stuck. Take a out of the first two and b out of the last two, and the same bracket appears twice.
Say the conclusion.
End with "therefore LHS > RHS". Showing the difference is positive and stopping there is working, not a finished proof.
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