How to prove Inequalities. Techniques to help prove that a < b | by Slawomir Chodnicki | Towards Data Science

When you develop an interest in mathematics and start working with material that takes proof seriously, sooner or later you’ll be trying to find proofs of inequalities.
Proving that something is equal to something else is usually somewhat easier. You manipulate both sides in the same manner until you arrive at the equation in question. You may have to make some clever substitutions, but there are only so many things you can do.
Inequalities can be a bit trickier because of transitivity. If you’re looking to show for some a and b that a < b, it may look very difficult. If no obvious solution presents itself, it may be necessary to come up with some c, for which you can show a < c < b, thus implying a < b.
Let’s look at examples of both cases, we’ll start off with a “simple” case for which we don’t need to invent intermediates.
All numbers in the examples are real numbers.
Given real numbers r and s, show the following:
Let’s first take the time to develop an intuition for what it says. Let’s pick some numbers for r and s and see what happens:
1/2 < 2/3 < 3/4 < 9/10 < 99/100
Alright, I see. I can get on board with that.
Now how do we go about proving that it’s always true?
In order to compare fractions, it helps to extend them so they have the same denominators. We’d have to extend the left side with (1+s) and the right side with (1+r)
Let’s do that, and see where we stand.
Now that the denominators are equal, it turns out we can get rid of them. We know that both (1+r) and (1+s) are positive numbers, so we know their product (1+r)(1+s) is a positive number. And so is the inverse 1/((1+r)(1+s)). We can multiply both sides by the inverse, effectively getting rid of the fractions.
We won’t have to flip the direction of the inequality since the number we’re multiplying with is known to be positive.
I’d say that’s progress. Let’s do it. We’re left with:
Multiplying the sums, we find that we end up with a common term on both sides: rs. We subtract it on both sides, arriving at a true statement as per our givens.
To present a formal proof, we just reverse our exploratory steps, so the reasoning makes sense when read top to bottom.
The proof was simple — in a sense — because it did not require us to get creative with any intermediate expressions. We could handle the proof very much like a proof of equality.
Let’s move on to something more demanding.
Given any two real numbers x and y , show that the following is
Oh boy. That one looks a bit more demanding. First, let’s get an intuition for what it says. Let’s pick x = y = 2, we get:
4/5 <= 2/3 + 2/3
Alright, let’s try another pair: x = 2, y = -3
-1/2 <= 2/3 + 3/4
OK, looks like we’re always going to stay below 1 on the left side, even go below 0. The right side can grow above 1, and will never go below 0 because the individual terms will never become negative.
Notice that when we pick x = 0 or y = 0 the left and right sides become equal.
Alright, We have a bit of an intuition for what’s going on. Let’s see if we can prove it.
The right-hand side has a sum of fractions. We can try forming the sum to see how it compares to the left-hand side. We need to expand the left term with 1+|y| and the right term with 1+|x|
Alright. So that’s what it looks like when we sum up the right-hand side. The inequality to prove becomes:
Proving inequalities, you often have to introduce one or more additional terms that fall between the two you’re already looking at. This often means taking away or adding something, such that a third term slides in.
Always check your textbook for inequalities you’re supposed to know and see if any of them seem useful. Exercises will often build upon each other too. Sometimes a result from exercise a) is key in exercise b). So don’t “forget” about your previous results.
In our case, we see that both sides look similar to our familiar pattern of something/(1+something) that we’ve just dealt with in our first proof.
Let’s see what it takes to get both sides to conform to our proven inequality, so we can use it.
Let’s look at our left-hand side.
If we take the absolute value of x+y of the numerator we could substitute r=|x+y| and our left-hand side would be of the form r/(1+r). Doing that would certainly grow the term. The absolute value of x+y is either greater — in case x+y<0 — or equal — in case x+y≥0.
So we have:
Looks like we’re done on the left side.
Let’s look at our right hand side.
What do we have to do to arrive at our desired form of s/(1+s)?
If we substitute s=|x|+|y|+|x||y|, we end up with (s+|x||y|)/(1+s). There’s one extra |x||y| present in our numerator. If we drop it, our term either stays the same —that would be the case if either x or y are 0 — or it shrinks. Because if neither x nor y are 0 then |x||y| is positive so we’re reducing the numerator in value.
Therefore we have:
And we’ve arrived at our desired form of s/(1+s)
We have transformed both sides to r/(r+1) and s/(s+1) form respectively. These are the terms we want to slide in between our initial left and right hand side.
If we can show that 0 ≤ r < s, we can use our first result of
r/(r+1) < s/(s+1)
We’ve substituted r=|x+y| and s=|x|+|y|+|x||y|. We see immediately that 0≤r, because it’s an absolute value. Through the triangle inequality we know |x+y|≤|x|+|y| is true, so we can show:
Great! We know 0 ≤ r ≤ s and can use r/(r+1) ≤ s/(s+1). We can put all our terms in order.
We’ve done it!
Our initial left side and right side terms were difficult to compare directly. But we could fit some terms between them that we knew how to compare, thus forming a chain that proves our initial statement.
When proving inequalities, it’s useful to look for ways to shrink or grow terms in a controlled way, such that they conform to known inequalities and form transitive chains.
If you feel like you have this down, why not test yourself and try and prove both inequalities from scratch? You’ve just read the solutions, so ... how hard can it be?
Here they are again.
First inequality:
Second inequality:
You got this!