Result 3.3.1.
Let \(x,y\in \mathbb{R}\text{.}\) Then \(x^2+y^2 \geq 2xy\text{.}\)
Section 3.3 Proofs of inequalities
Not all mathematics involves integers, nor do all proofs involve equalities. So we should do a few examples of inequalities involving real numbers. This isn’t just for variety, but does illustrate an important point about how our scratch work can often be quite different in logical structure from the final proof.
As always we start with some scratch work. We don’t know too many facts about inequalities — well, we do, but we haven’t stated too many of them as facts or axioms in this text. We do know that \(x^2 \geq 0\) no matter which real number we take for \(x\) — this was stated as Fact 3.0.2. So it would suffice to rearrange our inequality to make it look like the square of something.
\begin{equation*}
x^2 + y^2 - 2xy \geq 0
\end{equation*}
But this is precisely \((x-y)^2 \geq 0\text{,}\) which follows from the fact that we are squaring something.
So we see a way to prove things. But we should be very careful of the logical order here — how does the truth flow from one statement to another. Look at the structure of what we have done above.
- We started from the conclusion \(x^2+y^2 \geq 2xy\)
- We finished at the fact that the square of any real number is non-negative.
This order is not correct. We know that we must finish at the conclusion, not start at it. But we can reorder our work to give it the correct logical flow:
- Start from the fact that the square of a real number is non-negative.
- \((x-y)\) is a real, so its square is non-negative.
- Expand this expression and rearrange it
- Arrive at the conclusion.
This is quite common when we prove inequalities; the logical flow in scratch work is often the reverse of what is required for the proof. We typically start at the inequality we want to prove and then work our way to something we know — a fact, an axiom, a previous result or theorem. To then present the proof we must start at the axiom, fact or theorem, and then work our way to the result. We can now write things up nicely:
Proof.
Let \(x,y\) be real numbers. Hence \((x-y)^2 \geq 0\text{.}\) Expanding this gives \(x^2-2xy+y^2 \geq 0\text{.}\) This can then be rewritten as \(x^2+y^2 \geq 2xy\) and so gives the required result.
That was very illustrative (which is, of course, why we include this topic). Our scratch work can look very different from the final write up of the proof. We should do a couple more, but first a useful fact about inequalities, multiplication and division.
Fact 3.3.2.
Let \(a,b,c \in \mathbb{R}\) with \(a \geq b\text{.}\)
- If \(c \gt 0\) then \(ac \geq bc\) and \(a/c \geq b/c\text{.}\)
- If \(c \lt 0\) then \(ac \leq bc\) and \(a/c \leq b/c\text{.}\)
In proofs we will often need to combine inequalities together to make new inequalities. Very frequently we will make use of the fact that if \(a \gt b\) and \(b \gt c\) then we know that \(a \gt
c\text{.}\) This is the transitivity of “\(\gt\)” (see Section 9.2). For example, if we know \(a \gt
b \gt 0\) and \(c \gt d \gt 0\text{,}\) then by multiplying the first inequality by \(c\) we get \(ac \gt bc\text{.}\) Similarly, multiplying the second inequality by \(b\) we get \(bc \gt bd
\text{.}\) These two inequalities together imply that \(ac \gt
bd\text{.}\)
Result 3.3.3.
Let \(x \in \mathbb{R}\text{.}\) If \(x \geq 4\) then \(x^2-3x+7 \geq 11\text{.}\)
This one isn’t too bad and we should break things into pieces.
- We know that \(x \geq 4 \gt 0\text{,}\) so then multiplying this by \(x\) gives us \(x^2 \geq 4x\text{,}\) and similarly, multiplying it by \(4\) gives us \(4x \gt 16\text{.}\) Hence we know that \(x^2 \geq 16\text{.}\)
- Similarly, since we know \(x \geq 4\) we know that \(3x \geq 12\text{.}\) Ah —now there is a problem — we are about to try to take the difference of inequalities. Bad.
- Instead, go back and write \(x^2-3x = x(x-3)\text{.}\) Then since \(x \geq 4, (x-3) \geq 1\text{.}\) Hence \(x^2-3x=x(x-3) \geq x\text{.}\) So, because \(x \geq 4\) we know that \(x^2 -3x \geq 4\text{.}\)
- Adding \(7\) to both sides then gives us \(x^2-3x+7 \geq 4+7 = 11\text{.}\)
We should now carefully check the flow of logic. We do indeed start with the hypothesis \(x \geq 4\) and arrive at the conclusion. The order is good! Time to write it up.
Proof.
Let \(x \geq 4\) be a real number. Then we know that \(x-3 \geq 1\text{,}\) and so \(x(x-3) \geq 4\text{.}\) Thus \(x(x-3)+7 = x^2-3x+7 \geq 11\) as required.
In this case the logical flow in our scratch work matched the flow required for the proof. This is different from the previous example. There is not a hard rule that holds for all results. We need to be able to look at our scratch work, see the logical flow and determine how to translate that into a correct proof.
At the end of Chapter 5 we’ll prove the triangle inequality. We can’t do this just yet as it requires requires the development of a bit little more logical machinery.
