# Proving an inequality

Suppose $a$ and $b$ are real numbers. Prove that if $a<b$ then $\frac{a+b}{2}<b$.

The 'solution' hints at adding $b$ to both sides of the inequality $a<b$, and $a+b<2b$ is as far as I've got (I don't know the reason for adding $b$ to both sides, either - or rather, why you'd think to add the $b$ to each side to help finish the proof, instead of doing something else) - I'd like to know what to try next.

Source of exercise: How To Prove it: A Structured Approach, Second Edition - Daniel J. Velleman.

Thank-you for the responses. Is the following solution sufficient proof, or is there something I'm missing?

Proof. Suppose $a$ and $b$ are real numbers, and $a<b$. Adding $b$ to both sides of the inequality $a<b$, we get $a+b<2b$. Subsequently, we can divide both sides by 2, to get $\frac{a+b}{2}<b$. Therefore, if $a<b$, it follows then that $\frac{a+b}{2}<b$.

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Here is the reasoning beyond the approach. You deal with fraction(s), so the first natural thing is to bring everything to the same denominator or get rid of the denominator. If you bring everything to the same denominator (you can instead multiply everything by the common denominator), the inequality you need to prove is $\frac{a+b}{2} < \frac{2b}{2}$. Look now to the numerators, and try to see how can you get it from the first inequality.. The numerator on the left side has an extra $b$ and the numerator on the right side has ...... –  N. S. Jul 14 '11 at 15:24
Your proof is perfect! +1 –  t.b. Jul 14 '11 at 15:41
One thing I'd like to add: You can "see" that the inequality holds by drawing the real line, marking the two points $a \lt b$ on it and noticing that $\frac{a+b}{2}$ is the midpoint of $a$ and $b$. Saying that $a \lt b$ is equivalent to saying that the point $a$ is to the left of $b$, hence the midpoint $\frac{a+b}{2}$ must be to the left of $b$ or, in other words, $\frac{a+b}{2} \lt b$. –  t.b. Jul 14 '11 at 15:48
That's helpful, and interesting. Thanks! –  Daniel May Jul 14 '11 at 16:01

The reason for adding $b$ to both sides is that you want to get $a+b$. Once you've got $a+b < 2b$, you can divide both sides by $2$. The reason for doing that is that you want $(a+b)/2$. Here it matters that $2$ is positive: if you divide both sides by a negative number, then you'd have to change $<$ to $>$, but since $2$ is positive, $<$ remains $<$.

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We have $a < b$, but we want $a+b$ on the left hand side. So, an easy way to go from $a$ to $a+b$ on the left hand side is to add $b$ to the left hand side, and since we can't just add things to one side of the inequality, we better add $b$ to both sides. At this point, we're just trying things; there's no epiphany that "Oh, of course this is going to work!", just poking at the problem and seeing what we can do with it.

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Your proof is faultless. Of course you are assuming that adding the same thing to both sides preserves order, and that dividing both sides by a positive number also preserves order. However, these are by now to be considered standard facts in the course. Perhaps you should have said "Since $2$ is positive, we can $\dots$". But that may be unnecessarily fussy.

The "trick" of adding $b$ to both sides is fairly natural. It only has a faint whiff of magic.

Maybe you did the following before writing down a proof, at least I hope you did. On a "number line", put a dot for $a$, one for $b$, and one for $(a+b)/2$, halfway between them. This finishes things, you now know the result is true. It only remains to write down the details, using the notations and tools of your course. (Theo Buehler writes about the geometry in a comment which is far more important than the question that led to it.)

The geometry suggests the following alternative approach. Recall that $x<y$ iff $y-x>0$. So it is enough to prove that $b-\frac{a+b}{2}>0$. But we have $$b-\frac{a+b}{2}=\frac{2b-(a+b)}{2}=\frac{b-a}{2}>0$$ (the inequality $\frac{b-a}{2}>0$ follows from $a<b$.)

A little bit more complicated, but it arises from an understanding of what's happening underneath. And it tells you more than the simple fact that $b>(a+b)/2$. It tells you by how much $b$ is bigger than $(a+b)/2$.

By the way, for complicated expressions $X$ and $Y$, one of the standard strategies for showing that $X<Y$ is to show that $Y-X>0$.

Comment: Back to your solution. At the manipulational level, maybe this is how I would think. I want to show that $\frac{a+b}{2}<b$. Fractions are unpleasant, they are "broken" numbers (that's the correct etymology). So let's unbreak the fracture.

Note that the desired inequality holds iff $a+b <2b$. And $a+b<2b$ is an obvious consequence of $a<b$. Now we can hide the reasoning that led to the solution by writing the proof backwards. Or not.

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Here is how I would discover that one needs to add, or rather subtract, $\;b\;$, without any 'magic': I would start at the most complex side, and simplify from there, as follows:

\begin{align} & \tfrac {a+b} 2 < b \\ \equiv & \;\;\;\;\;\text{"arithmetic: multiply both sides by 2 -- to simplify the left hand side"} \\ & a+b < 2 \times b \\ \equiv & \;\;\;\;\;\text{"subtract $\;b\;$ from both sides -- to simplify both sides at the same time"} \\ & a < b \\ \end{align}

Now, some call this proof 'backwards'. However, simplifying $\;\tfrac {a+b} 2 < b\;$ is a lot easier than trying to add information to $\;a < b\;$: in the above calculation, at each point there is not much choice about what to do next, while $\;a < b\;$ is a starting point which allows too much choice.
Also, note how the above does not only prove the required "if $\;a < b\;$ then $\;\tfrac {a+b} 2 < b\;$", but it proves the stronger equivalence $$\tfrac {a+b} 2 < b \;\equiv\; a < b$$ for any real $\;a,b\;$.