Chapter 1 · Algebraic Proofs · Concept 5
1-15 Proving Inequalities (3)
Try the subtraction method on √a + √b and √(a + b). You get a difference full of roots that will not factorise and will not tell you its sign. The method that worked twice has run out. So we change the move.
Guiding question
How do you compare two numbers when both are hiding under roots?
Learning outcome. Prove inequalities involving square roots by squaring both sides.
Squaring keeps the order — for lengths
if x ≥ 0 and y ≥ 0, then x > y ⟺ x2 > y2
The arrow points both ways. So to compare two roots, compare their squares instead — and the squares have no roots left in them.
The words “if x ≥ 0 and y ≥ 0” are not decoration.
Look what happens without them. Start with −3 > −5, which is true. Square both sides: 9 > 25, which is false. Squaring flipped the answer because the numbers were negative.
Square roots are never negative, so you are safe here. But say it in your proof — write “both sides are non-negative” before you square. That sentence earns a mark.
The blue bar is the left side, the grey bar is the right. Move a and b anywhere you like, then press the button. The bars get taller — but the same one stays on top.
Both bars are lengths, so neither can be negative. Press the button and watch: squaring changes the numbers but never changes which bar is longer.
You will never make these two equal (unless one of a, b is 0) — that is why this one is >.
1. If a ≥ 0 and b ≥ 0, prove that √((a² + b²)/2) ≥ (a + b)/2, and say when they are equal.
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Say the safety line first.
both sides are ≥ 0, so squaring keeps the orderA root is never negative, and a + b is not negative because a and b are not. Write this down.
2. If a > 0 and b > 0, prove that √a + 4√b > √(a + 16b)
Work with a partner. Before either of you squares anything, both say out loud why it is allowed. Then square, and finish it the lesson 1-14 way.
Prove the following, where every letter stands for a non-negative number.
If a > b > 0, prove that √(a − b) < √a − b/(2√a).
Engineers use the right-hand side as a quick stand-in for the root on the left. This proves the shortcut always overshoots — so if it fits, the real thing fits.
Somebody proves −3 > −5 by squaring both sides and getting 9 > 25 — then says the original must be false. Where exactly did they go wrong?
Which condition did the Point! box insist on?
Squaring without saying why you may.
One sentence — "both sides are non-negative" — before the squaring. Without it the proof has a hole, and markers look for it.
(√a + √b)² is not a + b.
It is a + b + 2√(ab). The middle term is the whole reason these inequalities are true. Dropping it makes both sides equal and proves nothing.
Forgetting to square the bottom of a fraction.
((a + b)/2)² is (a + b)²/4, not (a + b)²/2. Half the lost marks on this page are that 4.
Choosing ≥ when it should be >.
Look at the gap between the squares. If it is a square like (√a − √b)², equality is possible and you write ≥. If it is something like 2√(ab) with a, b > 0, it can never be zero, so write >.
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