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)⁴ > = a³ +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 ...

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.

Since both sides of this are equal (by above) equality must have occurred at every stage of AM-GM. Therefore all the variables are equal, and as they add to n we have x₁=¼ = x_n=1.

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.)

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.

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?

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).

1/n

n å i=0

f(xi) ³ f(1/n

n å i=0

xi)

that is, the average of the function is greater than the function of the average. Try f(x)=x², 1/x, -logx, whatever else you feel like (but remember to check its second derivative first!).

David

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...

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:

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