Chapter 1 · Algebraic Proofs · Concept 5
1-14 Proving Inequalities (2)
This time nobody tells you that a is bigger than b. You are given nothing at all, and the inequality still has to be true for every number anyone puts in. So where can the positivity possibly come from?
Guiding question
What is always positive, no matter what number you use?
Learning outcome. Prove algebraic inequalities that hold for all real numbers, and state when the two sides are equal.
A square is never negative
(anything)2 ≥ 0
Put in a positive number and the square is positive. Put in a negative number and the square is still positive. The only way a square is zero is if the thing inside is zero.
So the method is the same as last lesson with one change:
work out LHS − RHS, then make it into squares
Once the difference is a square, or squares added together, you are done — there is nothing left to prove.
Drag the sliders as far as you like, in either direction. The green line is the gap between the two sides. It bends, it flattens, it touches the floor — but it never goes below.
The gap is LHS − RHS = (a − 2b)². Move the sliders. The green line is the gap, and it never crosses under zero.
Sometimes the difference is already a square and you just have to notice. Sometimes it is not, and you have to build one. That building is called completing the square.
The recipe, in one line:
x2 + bx → (x + b/2)2 − (b/2)2
Halve the number in front of x, put it in the bracket, then take away its square to keep the value the same. Example: x² − 6x → half of −6 is −3 → (x − 3)² − 9.
Check it if you like: (x − 3)² − 9 = x² − 6x + 9 − 9 = x² − 6x. ✓
1. Prove that a2 + 4b2 ≥ 4ab for all real a, b, and say when the two sides are equal.
- 1
Write LHS − RHS, exactly as last lesson.
a2 + 4b2 − 4abSame first line every time. Nothing crosses the sign.
2. Prove that x2 + y2 ≥ 2(x − y − 1)
Work with a partner. One of you writes the difference, the other hunts for the square. Then swap. Always state the equality case — it is worth marks.
Prove each inequality for all real values, and state when the sides are equal (except for 5 — look at that one and you will see why).
Prove that x² + 2y² − 10x + 12y + 43 ≥ 0 for all real x and y, and find the smallest value the left side can be.
Exercise 5 is x² + 2 > 2x, with a strict > sign. Every other question on the page uses ≥. What is it about the difference that decides which sign the question gets?
Work out the difference and look at whether it can ever be exactly zero.
Forgetting the equality case.
The book asks "when are they equal" almost every time. Set the inside of each square to zero and solve. If there are two squares, both must be zero together.
Halving before taking the number out.
For 2y² + 12y, take the 2 out first: 2(y² + 6y). Halving the 12 while the 2 is still there gives (y + 6)², which is wrong.
Not checking the middle term.
a² + 4b² − 4ab looks like (a − 2b)². Test it: 2 × a × 2b = 4ab ✓. If the middle term does not match, your square is not the right one.
"It is positive because I tried some numbers."
Trying numbers is not a proof — there are infinitely many you did not try. Only the square form proves it for every value at once.
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