You are currently browsing the category archive for the ‘investigation’ category.

This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).

**How to Avoid The Troll: A Puzzle**

In these situations it’s best to look at the extreme case first so you get some idea of the problem. If you are feeling particularly pessimistic you could assume that the troll is always going to be there. Therefore you would head to the top of the barrier each time. This situation is represented below:

**The Pessimistic Solution:**

Another basic strategy would be the optimistic strategy. Basically head in a straight line hoping that the troll is not there. If it’s not, then the journey is only 2km. If it is then you have to make a lengthy detour. This situation is shown below:

**The Optimistic Solution:**

The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.

The question is now, is there a better strategy than either of these? An obvious possibility is heading for the point halfway along where the barrier might be. This would make a triangle of base 1 and height 1/2. This has a hypotenuse of root (5/4). In the best case scenario we would then have a total distance of 2 x root (5/4). In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2. We find the expected value by multiply both by 0.5 and adding. This gives 2.63 (2 dp). But can we do any better? Yes – by using some algebra and then optimising to find a minimum.

**The Optimisation Solution:**

To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum. Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us. Wolfram Alpha is incredibly powerful -and also very easy to use. Here is what I entered:

and here is the output:

So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.

So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint. Not a huge difference – but nevertheless we’ll save ourselves a few seconds!

This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across? Maybe the troll appears only 1/3 of the time? Could you even generalise the results for when the troll is y distance away or appears z percent of the time?

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

**The Telephone Numbers – Graph Theory**

The telephone numbers are the following sequence:

1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496…

(where we start from n=0).

This pattern describes the total number of ways which a telephone exchange with n telephones can place a connection between pairs of people.

To illustrate this idea, the graph below is for n=4. This is when we have 10 telephones:

Each red line represents a connection. So the first diagram is for when we have no connections (this is counted in our sequence). The next five diagrams all show a single connection between a pair of phones. The last three diagrams show how we could have 2 pairs of telephones connected at the same time. Therefore the 4th telephone number is 10. These numbers get very large, very quickly.

**Finding a recursive formula**

The formula is given by the recursive relationship:

**T(n) = T(n-1) + (n-1)T(n-2)**

This means that to find (say) the 5th telephone number we do the following:

**T(5) = T(5-1) + (5-1)T(5-2)**

**T(5) = T(4) + (4)T(3)**

**T(5) = 10 + (4)4**

**T(5) = 26**

This is a quick way to work out the next term, as long as we have already calculated the previous terms.

**Finding an nth term formula
**

The telephone numbers can be calculated using the nth term formula:

This is going to be pretty hard to derive! I suppose the first step would start by working out the total number of connections possible between n phones – and this will be the the same as the graphs below:

These clearly follow the same pattern as the triangular numbers which is 0.5(n² +n) when we start with n = 1. We can also think of this as n choose 2 – because this gives us all the ways of linking 2 telephones from n possibilities. Therefore n choose 2 also generates the triangular numbers.

But then you would have to work out all the permutations which were allowed – not easy!

Anyway, as an example of how to use the formula to calculate the telephone numbers, say we wanted to find the 5th number:

We have n = 5. The summation will be from k = 0 and k = 2 (as 5/2 is not an integer).

Therefore T(5) = 5!/(2^{0}(5-0)!0!) + 5!/(2^{1}(5-2)!1!) + 5!/(2^{2}(5-4)!2!)

T(5) = 1 + 10 + 15 = 26.

**Finding telephone numbers through calculus**

Interestingly we can also find the telephone numbers by using the function:

y = e^{0.5x2+x}

and the nth telephone number (starting from n = 1) is given by the nth derivative when x = 0.

For example,

So when x = 0, the third derivative is 4. Therefore the 3rd telephone number is 4.

The fifth derivative of the function is:

So, when x =0 the fifth derivative is 26. Therefore the 5th telephone number is 26.

If you liked this post you might also like:

Fermat’s Theorem on the Sum of two Squares – A lesser known theorem from Fermat – but an excellent introduction to the idea of proof.

Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct. How?

**Happy Numbers**

Happy numbers are defined by the rule that you start with any positive integer, square each of the digits then add them together. Now do the same with the new number. Happy numbers will eventually spiral down to a number of 1. Numbers that don’t eventually reach 1 are called unhappy numbers.

As an example, say we start with the number 23. Next we do 2²+3² = 13. Now, 1²+3² = 10. Now 1²+o² = 1. 23 is therefore a happy number.

There are many things to investigate. What are the happy numbers less than 100? Is there a rule which dictates which numbers are happy? Are there consecutive happy numbers? How about prime happy numbers? Can you find the infinite cycle of sadness?

Nrich has a discussion on some of the maths behind happy numbers. You can use an online tool to test if numbers are happy or sad.

Perfect numbers are numbers whose proper factors (factors excluding the number itself) add to the number. This is easier to see with an example.

6 is a perfect number because its proper factors are 1,2,3 and 1+2+3 = 6

8 is not a perfect number because its proper factors are 1,2,4 and 1+2+4 = 7

Perfect numbers have been known about for about 2000 years – however they are exceptionally rare. The first 4 perfect numbers are 6, 28, 496, 8128. These were all known to the Greeks. The next perfect number wasn’t discovered until around 1500 years later – and not surprisingly as it’s 33,550,336.

The next perfect numbers are:

8,589,869,056 (discovered by Italian mathematician Cataldi in 1588)

137,438,691,328 (also discovered by Cataldi)

2,305,843,008,139,952,128 (discovered by Euler in 1772).

and they keep getting bigger. The next number to be discovered has 37 digits are was discovered over 100 years later. Today, even with vast computational power, only a total of 48 perfect numbers are known. The largest has 34,850,340 digits.

There are a number of outstanding questions about perfect numbers. Are there an infinite number of perfect numbers? Is there any odd perfect number?

Euclid in around 300BC proved that that 2^{p−1}(2^{p}−1) is an even perfect number whenever 2^{p}−1 is prime. Euler (a rival with Euclid for one of the greatest mathematicians of all time), working on the same problem about 2000 years later went further and proved that this formula will provide *every* even perfect number.

This links perfect numbers with the search for Mersenne Primes – which are primes in the form 2^{p}−1. These are themselves very rare, but every new Mersenne Prime will also yield a new perfect number.

The first Mersenne Primes are

(2^{2}−1) = 3

(2^{3}−1) = 7

(2^{5}−1) = 31

(2^{7}−1) = 127

Therefore the first even perfect numbers are:

2^{1}(2^{2}−1) = 6

2^{2}(2^{3}−1) = 28

2^{4}(2^{5}−1) = 496

2^{6}(2^{7}−1) = 8128

**Friendly Numbers**

Friendly numbers are numbers which share a relationship with other numbers. They require the use of σ(a) which is called the divisor function and means the addition of all the factors of a. For example σ(7) = 1 + 7 = 8 and σ(10) = 1 +2 +5 + 10 = 18.

Friendly numbers therefore satisfy:

σ(a)/a = σ(b)/b

As an example (from Wikipedia)

σ(6) / 6 = (1+2+3+6) / 6 = 2,

σ(28) / 28 = (1+2+4+7+14+28) / 28 = 2

σ(496)/496 = (1+2+4+8+16+31+62+124+248+496)/496 = 2

Therefore 28 and 6 are friendly numbers because they share a common relationship. In fact all perfect numbers share the same common relationship of 2. This is because of the definition of perfect numbers above!

Numbers who share the same common relationship are said to be in the same club. For example, 30,140, 2480, 6200 and 40640 are all in the same club – because they all share the same common relationship 12/5.

(eg. σ(30) /30 = (1+2+3+5+6+10+15+30) / 30 = 12/5 )

Are some clubs of numbers infinitely big? Which clubs share common integer relationships? There are still a number of unsolved problems for friendly numbers.

**Solitary Numbers**

Solitary numbers are numbers which don’t share a common relationship with any other numbers. All primes, and prime powers are solitary.

Additionally all number that satisfy the following relationship:

HCF of σ(a) and a = 1.

are solitary. All this equation means is that the highest common factor (HCF) of σ(a) and a is 1. For example lets choose the number 9.

σ(9)= 1+3+9 = 13. The HCF of 9 and 13 = 1. So 9 is solitary.

However there are some numbers which are not prime, prime powers or satisfy HCF (σ(a) and a) = 1, but which are still solitary. These numbers are much harder to find! For example it is believed that the following numbers are solitary:

10, 14, 15, 20, 22, 26, 33, 34, 38, 44, 46, 51, 54, 58, 62, 68, 69, 70, 72, 74, 76, 82, 86, 87, 88, 90, 91, 92, 94, 95, 99

But no-one has been able to prove it so far. Maybe you can!

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

This carries on the previous investigation into Farey sequences, and is again based on the current Nrich task Ford Circles. Below are the Farey sequences for F_{2}, F_{3} and F_{4}. You can read about Farey sequences in the previous post.

This time I’m going to explore the link between Farey sequences and circles. First we need the general equation for a circle:

This has centre (p,q) and radius r. Therefore

**Circle 1:**

has centre:

and radius:

**Circle 2:**

has centre:

and radius:

Now we can plot these circles in Geogebra – and look for the values of a,b,c,d which lead to the circles touching at a point.

**When a = 1, b = 2, c = 2, d = 3:**

Do we notice anything about the numbers a/b and c/d ? a/b = 1/2 and c/d = 2/3 ? These are consecutive terms in the F_{3 }sequence. So do other consecutive terms in the Farey sequence also generate circles touching at a point?

**a = 1, b = 1, c = 2, d = 3**

Again we can see that the fractions 1/1 and 2/3 are consecutive terms in the F_{3 }sequence. So by drawing some more circle we can graphically represent all the fractions in the F_{3 }sequence:

So these four circles represent the four non-zero fractions of in the F_{3 }sequence!

and this is the visual representation of the non-zero fractions of in the F_{4 }sequence. Amazing!

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

This is a mini investigation based on the current Nrich task Farey Sequences.

As Nrich explains:

I’m going to look at Farey sequences (though I won’t worry about rearranging them in order of size). Here are some of the first Farey sequences. The missing fractions are all ones which simplify to a fraction already on the list (e.g. 2/4 is missing because this is the same as 1/2)

You should be able to notice that the next Farey sequence always contains the previous Farey sequence, so the problem becomes working out which of the new fractions added will not cancel down to something already on the list.

**Highest Common Factors**

Fractions will not cancel down (simplify) if the numerator and denominator have a highest common factor (HCF) of 1. For example 2/4 simplifies because the highest common factor of 2 and 4 is 2. Therefore both top and bottom can be divided by 2. 4/5 does not simplify because the HCF of 4 and 5 is 1.

We call 2 numbers which have a HCF of 1 **relatively prime.**

for example for the number 4: 1 and 3 are both relatively prime (HCF of 1 and 4 =1, HCF of 3 and 4 = 1).

**Relatively prime numbers**

2: 1

3: 1,2

4: 1,3

5: 1,2,3,4

6: 1,5

7: 1,2,3,4,5,6

8: 1,3,5,7

9: 1,2,4,5,7,8

You might notice that these give the required numerators for any given denominator – i.e when the denominator is 9, we want a numerator of 1,2,4,5,7,8.

**Euler totient function**

Euler’s totient function is a really useful function in number theory – which counts the number of relatively prime numbers a given number has. For example from our list we can see that 9 has 6 relatively prime numbers.

Euler’s totient function is defined above – it’s not as complicated as it looks! The strange symbol on the right hand side is the product formula – i.e we multiply terms together. It’s easiest to understand with some examples. To find Euler’s totient function we first work out the prime factors of a number. Say we have the number 8. The prime factors of 8 are 2^{3}. Therefore the only unique prime factor is 2.

Therefore the Euler totient function tells me to simply do 8 (1 – 1/2) = 4. This is how many relatively prime numbers 8 has.

Let’s look at another example – this time for the number 10. 10 has the prime factorisation 5 x 2. Therefore it has 2 unique primes, 2 and 5. Therefore the Euler totient function tells me to do 10(1-1/2)(1-1/5) = 4.

One more example, this time with the number 30. This has prime factorisation 2 x 3 x 5. This has unique prime factors 2,3,5 so I will do 30(1 -1/2)(1-1/3)(1-1/5) =8.

**An equation for the number of fractions in the Farey sequence**

Therefore I can now work out how many fractions will appear in a given Farey sequence. I notice that for (say) F_{5} I will add Euler’s totient for n = 2, n = 3, n = 4 and n = 5. I then add 2 to account for 0/1 and 1/1. Therefore I have:

For example to find F_{6}

There are lots of things to investigate about Farey functions – could you prove why all Farey sequences have an odd number of terms? You can also look at how well the Farey sequence is approximated by the following equation:

For example when n = 10 this gives:

and when n = 1000 this gives:

These results compare reasonably well as an estimation to the real answers of 33 and 304,193 respectively.

**IB Revision**

**Circular Motion: Modelling a ferris wheel**

This is a nice simple example of how the Tracker software can be used to demonstrate the circular motion of a Ferris wheel. This is sometimes asked in IB maths exams – so it’s nice to get a visual representation of what is happening.

First I took a video from youtube of a Ferris wheel, loaded it into Tracker, and then used the program to track the position of a single carriage as it moved around the circle. I then used Tracker’s graphing capabilities to plot the height of the carriage (y) against time (t). This produces the following graph:

As we can see this is a pretty good fit for a sine curve. So let’s use the regression tool to find what curve fits this:

The pink curve with the equation:

y = -116.1sin(0.6718t+2.19)

fits reasonably well. If we had the original dimensions of the wheel we could scale this so the y scale represented the metres off the ground of the carriage.

There we go! Short and simple, but a nice starting point for an investigation on circular motion.

**IB Revision**

**Project Euler: Coding to Solve Maths Problems**

Project Euler, named after one of the greatest mathematicians of all time, has been designed to bring together the twin disciplines of mathematics and coding. Computers are now become ever more integral in the field of mathematics – and now creative coding can be a method of solving mathematics problems just as much as creative mathematics has always been.

The first problem on the Project Euler Page is as follows:

*If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.*

*Find the sum of all the multiples of 3 or 5 below 1000.*

This is a reasonably straight forward maths problem which we can solve using the summation of arithmetic sequences (I’ll solve it below!) but more interestingly is how a computer code can be written to solve this same problem. Given that I am something of a coding novice, I went to the Project Nayuki website which has an archive of solutions. Here is a slightly modified version of the solution given on Project Nayki, designed to run in JAVA:

The original file can be copied from here, I then pasted this into an online JAVA site jdoodle. The only modification necessary was to replace:

*public final class p001 implements EulerSolution* with *public class p001*

Then after hitting execute you get the following result:

i.e the solution is returned as 233,168. Amazing!

But before we get carried away, let’s check the answer using some more old fashioned maths. We can break down the question into simply trying to find the sum of multiples of 3 under 1000, the sum of the multiples of 5 under 1000 and then subtracting the multiples of 15 under 1000 (as these will have been double counted). i.e:

(3 + 6 + 9 + … 999) + (5 + 10 + 15 + … 995) – (15 + 30 + 45 + …990)

This gives:

S_333 = 333/2 (2(3)+ 332(3)) = 166,833

+

S_199 = 199/2 (2(5) + 198(5)) = 99, 500

–

S_66 = 66/2 (2(15) +65(15) = 33, 165.

166,833 + 99, 500 – 33, 165 = 233, 168 as required.

Now that we have seen that this works we can modify the original code. For example if we replace:

if (i % 3 == 0 || i % 3 == 0)

with

if (i % 5 == 0 || i % 7 == 0)

This will find the sum of all the multiples of 5 or 7 below 1000. Which returns the answer 156,361.

Replacing the same line with:

if (i % 5 == 0 || i % 7 == 0 || i % 3 == 0)

will find the sum of all the multiples of 3 or 5 or 7 below 1000, which returns the answer 271,066. To find this using the previous method we would have to do:

Sum of 3s + Sum of 5s – Sum of 15s + Sum of 7s – Sum of 21s – Sum 35s – Sum of 105s. Which starts to show why using a computer makes life easier.

This would be a nice addition to any investigation on Number Theory – or indeed a good project for anyone interested in Computer Science as a possible future career.

**Measuring the Distance to the Stars**

This is a very nice example of some very simple mathematics achieving something which for centuries appeared impossible – measuring the distance to the stars. Before we start we need a few definitions:

- 1 Astronomical Unit (AU) is the average distance from the Sun to the Earth. This is around 150,000,000km.
- 1 Light Year is the distance that light travels in one year. This is around 9,500,000,000,000km. We have around 63000AU = 1 Light Year.
- 1 arc second is measurement for very small angles and is 1/3600 of one degree.
- Parallax is the angular difference in measurement when viewing an object from different locations. In astronomy parallax is used to mean the half the angle formed when a star is viewed from opposite sides of the Earth’s solar orbit (marked on the diagram below).

With those definitions it is easy to then find the distance to stars. The parallax method requires that you take a measurement of the angle to a given star, and then wait until 6 months later and take the same measurement. The two angles will be slightly different – divide this difference by 2 and you have the parallax.

Let’s take 61 Cyngi – which Friedrick Bessel first used this method on in the early 1800s. This has a parallax of 287/1000 arc seconds. This is equivalent to 287/1000 x 1/3600 degree or approximately 0.000080 degrees. So now we can simply use trigonometry – we have a right angled triangle with opposite side = 1 AU and angle = 0.0000080. Therefore the distance is given by:

tanΦ = opp/adj

tan(0.000080) = 1/d

d = 1/tan(0.000080)

d = 720000 AU

which is approximately 720000/63000 = 11 light years away.

That’s pretty incredible! Using this method and armed with nothing more than a telescope and knowledge of the Earth’s orbital diameter, astronomers were able to judge the distance of stars in faraway parts of the universe – indeed they used this method to prove that other galaxies apart from our own also existed.

**Orion’s Belt**

The constellation of Orion is one of the most striking in the Northern Hemisphere. It contains the “belt” of 3 stars in a line, along with the brightly shining Rigel and the red super giant Betelgeuse. The following 2 graphics are taken from the great student resource from the Royal Observatory Greenwich:

The angles marked in the picture are in arc seconds – so to convert them into degrees we need to multiply by 1/3600. For example, Betelgeuse the red giant has a parallax of 0.0051 x 1/3600 = 0.0000014 (2sf) degrees. Therefore the distance to Betelgeuse is:

tanΦ = opp/adj

tan(0.0000014) = 1/d

d = 1/tan(0.0000014)

d = 41,000,000 AU

which is approximately 41,000,000/63000 = 651 light years away. If we were more accurate with our rounding we would get 643 light years. That means that when we look into the sky we are seeing Betelgeuse as it was 643 years ago.

**IB Revision**

This post is inspired by the Quora thread on interesting functions to plot.

**The butterfly**

This is a slightly simpler version of the butterfly curve which is plotted using polar coordinates on Desmos as:

Polar coordinates are an alternative way of plotting functions – and are explored a little in HL Maths when looking at complex numbers. The theta value specifies an angle of rotation measured anti-clockwise from the x axis, and the r value specifies the distance from the origin. So for example the polar coordinates (90 degrees, 1) would specify a point 90 degrees ant clockwise from the x axis and a distance 1 from the origin (i.e the point (0,1) in our usual Cartesian plane).

2. **Fermat’s Spiral**

This is plotted by the polar equation:

The next 3 were all created by my students.

3. **Chaotic spiral (by Laura Y9)**

I like how this graph grows ever more tangled as it coils in on itself. This was created by the polar equation:

4. **The flower (by Felix Y9)**

Some nice rotational symmetries on this one. Plotted by:

5. **The heart (by Tiffany Y9)**

Simple but effective! This was plotted using the usual x,y coordinates:

You can also explore how to draw the Superman and Batman logos using Wolfram Alpha here.

**A geometric proof for the Arithmetic and Geometric Mean**

There is more than one way to define the mean of a number. The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:

(x_{1}.x_{2}.x_{3}…x_{n})^{1/n}

So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)^{1/3}.

With 2 numbers, a and b, the geometric mean is (ab)^{1/2}.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)^{1/2} for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems). Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle. Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras: (MN)^{2} = a^{2} +x^{2}

We find NQ using Pythagoras: (NQ)^{2} = b^{2} +x^{2}

Therefore the length MQ can also be found by Pythagoras:

(MQ)^{2} = (MN)^{2 } + (NQ)^{2}

(MQ)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

But MQ = a + b. Therefore:

(a + b)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

a^{2}+ b^{2} + 2ab = a^{2} +x^{2} + b^{2} +x^{2}

2ab = x^{2} +x^{2}

ab = x^{2}

x = (ab)^{1/2}

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)^{1/2}. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.