Season 3 Episode 5
In this episode, Ittihad introduces inequalities, from AM-GM to the Power Mean Inequality.
Further Reading
AM-GM
Here’s a statement of the AM-GM inequality again.
For positive numbers $a_1$, $a_2$,$\dots$, $a_n$, we have
\begin{equation*}
\frac{a_1+a_2+\dots +a_n}{n}\geq \sqrt[n]{a_1\times a_2\times \dots \times a_n}
\end{equation*}
The expression on the left is called the arithmetic mean (AM), and the expression on the right is the geometric mean (GM).
Average Problems
Exercise: Three positive numbers $a$, $b$, $c$ are in an arithmetic progression (common difference between terms). Find $b$ in terms of $a$ and $c$.
Exercise: Three positive numbers $a$, $b$, $c$ are in a geometric progression (common ratio between terms). Find $b$ in terms of $a$ and $c$.
Exercise: If a cuboid has side lengths $a$, $b$, $c$, and it has the same volume as a cube. Find the side length of the cube in terms of $a$ and $b$ and $c$.
Exercise: Three circles are each tangent to the same two lines, with one circle in the middle also tangent to the other two circles. Given the radius of the outside circles are 1 and 20, what's the radius of the middle circle?
A version of this question appeared in MAT 2016 Q4, but people prefer it if I don't talk about that test at this club!
Proof of AM-GM
Ittihad mentioned a proof of the AM-GM inequality by "Forwards-Backwards Induction". It's a really cool proof, and rather than reproducing it here I'm going to link to the proof on the Art of Problem Solving website.
AoPS is a nice resource, but be warned; it has a very higher upper bound on the difficulty of its articles. It's very simple to get from this resource to an incredibly difficult one in just a couple of clicks! If you're using a screen reader, then note that the way AoPS presents mathematics is different from the way this site presents mathematics; they've put LaTeX code into image alt-text fields.
The AoPS page on the AM-GM inequality includes some exercises at the end of the page. The introductory and intermediate problems seem appropriate to me.
Muirhead's Inequality
Ittihad also mentioned something called Muirhead's inequality. This is quite hard to explain in general, so I'll start with an example of the sort of thing it can be used to prove.
In the livestream we used the AM-GM inequality to prove things like
\begin{equation*}
a^4+b^4+c^4\geq a^2bc+b^2ac+c^2ab
\end{equation*}
where on the right-hand side, the pattern of powers is the same for each term, but the letters cycle around through the different permutations. We could also think of the left-hand side as something similar, if we write the terms as
\begin{equation*}
a^4b^0c^0+b^4a^0c^0+c^4a^0b^0\geq a^2bc+b^2ac+c^2ab
\end{equation*}
remembering that $x^0=1$ for any positive $x$. So this inequality is a sort of comparison between the set of powers $(4,0,0)$ and the set of powers $(2,1,1)$, and the first set of powers "wins".
Muirhead's inequality is all about comparing expressions like this, and gives us a way to decide which set of powers will win.
The result is a bit fiddly, and I need to define a new word. Given two sequences of powers like this; not increasing, and the same length, with the same sum, we say that the first sequence of powers "majorizes" the second if the sum of the first $k$ terms in the first sequence is always bigger than or equal to the sum of the first $k$ terms of the second sequence.
That's a lot to take in! To check that $(4,0,0)$ majorizes $(2,1,1)$, we need to check that
- the length of these sequences is the same (yes, they both have three terms)
- the sum is the same (yes, both sum to 4)
- ($k=1$) the first term of the first sequence is bigger than or equal to the first term of the second sequence (yes, $4\geq2$)
- ($k=2$) the sum of the first two terms is bigger or equal for the first sequence (yes, $4+0\geq 2+1$)
One way to think about this is to imagine that the sequences represent points scored by two teams in different rounds of a competition. If the second team is never ahead at any stage, then the first sequence majorizes the second sequence.
OK, here we go with Muirhead's inequality.
Muirhead's inequality states that if the sequence of powers $(p_1,p_2,p_3,\dots,p_n)$ majorizes the sequence of powers $(q_1,q_2,q_3,\dots q_n)$, and if $a_1,a_2, \ldots, a_n$ are positive real numbers then
\begin{equation*}
\sum_{sym}a_1^{p_1}a_2^{p_2}\ldots a_n^{p_n} \geq \sum_{sym}a_1^{q_1}a_2^{q_2}\ldots a_n^{q_n}
\end{equation*}
where $\displaystyle \sum_{sym}$ ("the symmetric sum") means that we go through all the permutations of $a_1,a_2,a_3,\dots,a_n$ and add all those terms together.
For example, the sequence $(6,1,0)$ majorizes the sequence $(5,1,1)$. Muirhead's inequality gives us
\begin{equation*}
a^6b^1c^0+a^6c^1b^0+b^6a^1c^0+b^6c^1a^0+c^6a^1b^0+c^6b^1a^0 \geq a^5b^1c^1+a^5c^1b^1+b^5a^1c^1+b^5c^1a^1+c^5a^1b^1+c^5b^1a^1
\end{equation*}
Note that some of the terms on the right-hand side are equal, but the symmetric sum really does include all permutations (without worrying about whether the powers are the same or not). This simplifies a bit to
\begin{equation*}
a^6b+a^6c+b^6a+b^6c+c^6a+c^6b \geq 2\left(a^5bc+b^5ac+c^5ab\right)
\end{equation*}
The AoPS website again gives some examples and even a proof (which is quite technical this time).
Jensen’s Inequality
On the livestream we also talked about Jensen's Inequality, which is described in more detail here. The version on that page includes some "weights" on each term; in the livestream version of the inequality we took those to all be $\displaystyle\frac{1}{n}$ but Jensen's inequality still holds even if the weights are not equal. I'm calling them weights because I'm still thinking about centre of mass.
There's a proof on that webpage which is far beyond anything we've done on the Maths Club; it uses some maths from a typical first-year university course, and has some gnarly notation. You've been warned!
Don't be mean
There are some other averages that we should talk about!
Exercise: I walk to the shops at a speed of $3 \text{ms}^{-1}$ and then immediately run back at a speed of $5 \text{ms}^{-1}$ without buying anything (I have forgotten my wallet). What's my average speed? Hint: not $4 \text{ms}^{-1}$, think about total distance and total time.
The harmonic mean (HM) of $a$ and $b$ is defined as $\displaystyle\frac{1}{\frac{1}{a}+\frac{1}{b}}$.
Exercise: Prove that the harmonic mean is $\leq$ the geometric mean.
The quadratic mean (QM) or root mean square of $a$ and $b$ is $\displaystyle\sqrt{\frac{a^2+b^2}{2}}$. This is good for talking about average speeds of random velocity vectors, and it's also used as a measure of average error in some situations.
Exercise: Prove that the quadratic mean is $\geq$ the arithmetic mean.
If you like this sort of thing, then you might like the book Introduction to Inequalities by the UKMT. If you're lucky, then you might be able to persuade your school library to pick up a copy. The UKMT organise olympiad-style Mathematics competitions in the UK, as well as lots of other maths activities.
If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk.