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Inequalities solving an Equation
By Amars Birdi on Thursday, July 05, 2001 - 10:27 am:

Could anyone please show me if there are any non-obvious solutions to the following set of equations (or, alternatively, prove that the solution containing ones is the only solution):

a+b+c+d+ ('nth'letter) = n
a4+..+('nth' letter)4³a3+b3+..+('nth'letter) 3

I have tried to attack this problem in two basic ways, first I assumed that there were several solutions to the equations and I used the expansions for (a+b)2, (a+b)3 and (a+b)4 (for the n=2 case). However, I don't see any way of generalising the proof that the only solution for n=2 is when a=b=1. I then tried using the AMGM inequalities to prove that the only solution is a=b=c=...=1 - that turned out to be useless.
Any help for the solutions to this equation set would be liked.
Thanks ...

By David Loeffler on Thursday, July 05, 2001 - 05:09 pm:

Yes, the obvious solution with ones is the only solution in reals. You had the right idea when you tried inequalities; but you need a slightly different one.
Have you ever heard of Chebshev's inequality? It is a very useful result.

Consider two sequences ai and bi, where i runs from 1 to n. Then if (ai-aj)(bi-bj) ³ 0 for all i, j, then we have 1/n Sn i=1 aibi ³2(Sn i=1 ai)(Sn i=1 bi).
Equality holds if and only if one or both of the sequences is identically equal to a constant.

Can you see how to apply this to your problem?

By Michael Doré on Thursday, July 05, 2001 - 05:09 pm:

If I understand correctly, we have:

x1 + x2 + ... + xn = n
x14 + x24 + ... + xn4 = x13 + x23 + ... + xn3

where I've let x1 = a, x2 = b, etc.

You can now divide the second equation by the first equation, re-arrange, and you get:

n(x14 + ... + xn4) = (x13 + ... + xn3)(x1 + ... + xn)

Now the equation is homogeneous so is much easier to work with.

If you now multiply the RHS out, and collect fourth powers on the LHS, you end up with:

(n-1)(x14 + x24 + ... + xn4) = S i¹jxixj3

By AM-GM we have:

xixj3 £ xi4/4 + 3xj4/4

with equality iff xi = xj.

Add the inequality to itself for all distinct pairs (i,j) and you get:

S i ¹ jxixj3 £ (n-1)(x14 + ... + xn4)

Since both sides of this are equal (by above) equality must have occured at every stage of AM-GM. Therefore all the variables are equal, and as they add to n we have x1 = ... = xn = 1.

By Amars Birdi on Friday, July 06, 2001 - 11:04 am:

Sorry David, but I haven't seen Chebshev's inequality before, but I will attempt to apply it to the problem and see if it gives a slightly diffrent method of solution to Michaels solution.
Thanks for the hints and solutions!
(P.S - this was a problem fron the 'Komal' site that I had managed to dredge up from an old folder.)

By David Loeffler on Friday, July 06, 2001 - 10:26 pm:

I suppose a lot of inequalities like this have more than one solution and you can pretty much pick whichever one of the standard inequalites you think will work. Maybe the Chebyshev route is a bit roundabout; sorry. Anyway, have fun.

By Amars Birdi on Sunday, July 08, 2001 - 03:42 pm:

I have been thinking about both the Cheshev inequalitiy and the AMGM inequality. Are these meant to be particular cases of the power mean inequality? How does one prove this more general inequality?

By David Loeffler on Monday, July 09, 2001 - 08:37 pm:

Chebyshev's inequality isn't a case of the power mean inequality. AM-GM can be regarded as such, but only with a little fiddling around (the limit of the nth power mean as n->0 is the geometric mean, but this is quite fiddly to prove).

However, AM-GM and the power mean inequality are both cases of a wonderfully powerful result called Jensen's inequality. Define a function f(x) to be convex if f''(x) ³ 0 for all x. Then if f is convex, we have for all sequences xi (i runs from 1 to n) that:
1/n Sn i=0 f(xi) ³ f(1/n Sn i=0 xi)
that is, the average of the function is greater than the function of the average. Try f(x)=x2, 1/x, -log x, whatever else you feel like (but remember to check its second derivative first!).

David

By Amars Birdi on Friday, July 20, 2001 - 06:02 pm:

That seems like a very general result (it seems as if maths is full of such theorems which link apparently unlinkable mathematical topics). You wouldn't happen to tell me why the second derivative of f(x) being greater than zero for all x implies that the function is convex (from a graphical view what does being convex mean?). Also, where does this Chebshev inequality come from? I remember seeing this inequality in a text book to do with vectors, but I don't remember the proof of it. I would also appreciate it if somebody could point me in the general direction of possibly an inductive proof of the AMGM inequality and the set of equations that started this thread (I ask for an inductive proof for the set of equations as I attempted some kind of solution using this method when I first saw the problem!). Thanks...

By David Loeffler on Friday, July 20, 2001 - 10:52 pm:

Well, there are various definitions of convexity and associated deep theorems, but in this context we say a function is convex if the line connecting two points on its graph lies wholly above the graph. For example, f(x)=x2 is convex; picture the line connecting any two points on a parabola.

Intuitively, a function is convex if it "curves upwards."

As for Chebyshev's inequality, are you sure it was in a book on vectors? There are several inequalities that are effectively statements about vectors (principally the Cauchy-Schwartz inequality - have you heard of this?) but as far as I know there is no pure vector form of Chebyshev. Chebyshev's is actually a rather odd inequality and doesn't sit well with any others I know of.

The way I tend to think of Chebyshev's inequality is in terms of statistics. When you have a set of n points on a graph and you are fitting the best possible straight line to them, you get the following expression for the slope of the line:
(n Sn i=1xiyi - Sn i=1xiSn j=1yj)/(n Sn i=1xi2 - (Sn i=1xi)2)

Now when x and y are similarly ordered, so y is large when x is large, and small when x is small, then you would expect the line to have positive slope; and that is exactly what Chebyshev tells us, since the expression in the numerator is precisely that which occurs in Chebyshev when you recast it as something > 0 (multiplied by n2). So that is how I remember Chebyshev.

David

(PS. Let's kill this discussion now and continue in the new one at the top of the list.)

By Amars Birdi on Sunday, July 29, 2001 - 01:36 pm:

You are quite right David when you say that it was the Cauchy Scwarz inequality in my vectors book (not the Chebyshev one!).