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If z=100, then f(100)=f(f(111))=f(101)=91. If 90<=z<=99,
then f(z)=f(f(z+11))=f(z+1). Thus f(90)=f(91)= ... =f(99)=f(100)=91.
Let f(z)=91. Then f(z-11)=f(f(z))=f(91)=91 holds for all z<=100.
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We will prove after Alexej Cimadiev from Ukraine that any
function f(x) defined for all real x (i.e. also for
absolute value function) can be expressed as a sum of two central symmetric
functions f1(x) and f2(x). Let
the centres of symmetries are in points [1;0] for f1
and [0;0] for f2. Furthemore, let f1(x)=0
for all x from the interval [0;1]. Because f2(x)=f(x)-f1(x),
we have f2(x)=f(x) for all x from [0;1].
Thanks to the symmetry of f1 we get the value of both functions in
the interval [1;2]: f1(x)=0, f2(x)=f(x). Symmetry
of the function f2(x) determines both functions in the interval
[-2;0] - in the case of abslute value f1(x)=-2x and
f2(x)=-x. Next, the symmetry of f1 gives
the interval [2;4] and by continuing in this process we acquire the
the value for both functions for all real x.
It is evident from the construction that both functions will have central-symmetric
graphs and that f(x)=f1(x)+f2(x).
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Cauchy inequality for real numbers x1, x2, ... ,xk
and y1, y2, ..., yk states that:
( x1y1 + x2y2 + ... + xkyk)^2 <=
( x1^2 + x2^2 + ... + xk^2 ) (y1^2 + y2^2 + ... + yk^2)
and the left and right sides are equal when the vectors (x1,x2,...,xk)
and (y1,y2,...yk) are linearly dependent. Particularly,
for k=2, x1=ai, x2=1, y1=1,
y2=1/bi we get
(ai + 1/bi)^2 <= (ai^2 + 1)(1+1/bi^2)
This inequality holds for all i=1,2,...,n. This means that
(a1 + 1/b1)(a2 + 1/b2) ... (an + 1/bn) <=
<= sqrt( (a1^2 + 1)(1 + 1/b1^2)(a2^2 + 1)(1 + 1/b2^2)...(an^2 + 1)(1 + 1/bn^2) =
= sqrt( (a1^2 + 1)(a2^2 + 1) ... (an^2 + 1)(1 + 1/a1^2)(1 + 1/a2^2) ... (1 + 1/an^2)
Left-hand-side contains a constant independent from the permutation. But the equality sets in
for a1=b1, a2=b2, ..., an=bn,
because
(x + 1/x)^2 = (x^2 + 1)(1 + 1/(x^2)). Thus the maximum of our expression is the product
at the right-hand-side of the last inequality above.
Note: The task was to look for a maximum of the expression through all permutations, i.e.
numbers a1, a2, ... an were fixed.
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This problem seemed complicated and we have received two kinds of correct solutions.
This solution is after Martin Potocny and Mansur Boase:
First, we prove by contradiction that the function f(x) is injective
(i.e. if x<>y, then f(x)<>f(y)).
Let f(x)=f(y). Then x=f(f(f(x))) = f(f(f(y))) = y and that's the
contradiction.
So f(x) is injective and continuous. In addition, we prove that it is
also monotone (increasing or decreasing). In case it were not, there would be a point
x0 where the function would be changing from increasing to decreasing
(or vice versa), thus x0 would be an extreme. Points in the neighbourhood
of x0 would be all greater or all lower than f(x0).
In such case, there would be two points x0,1, x0,2 that
f(x0,1)=f(x0,2), but this is not possible, since f is
injective (Note: more precise mathematical proof requires more precise mathematical definition
of continuos function and it was not required).
Function f is monotone.
- Let f is decreasing and x < y are two arbitrary real numbers.
Then
f(x) > f (y); f(f(x)) < f(f(y)); x = f(f(f(x))) > f(f(f(y))) = y,
which is a contradiction.
- Let f is increasing. We consider these cases:
- If there exists such x that f(x)>x, then because f is increasing:
f(f(x)) > f(x) > x; x = f(f(f(x))) > f(f(x)) > f(x) > x,
which is a contradiction.
- If there exists such x that f(x) < x, then because f is increasing:
f(f(x)) < f(x) < x; x = f(f(f(x))) < f(f(x)) < f(x) < x,
which is again a contradiction.
That's why f(x)=x, for all x.
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First, we prove using mathematical induction that f is increasing.
- Since f(x) is a natural number for all x, there exists a natural
number m, that f(m) is the minimal. If m>1, then f(m)>f(f(m-1))
and thus f(m) would not be minimal, which is contradiction. That's why m=1
and f(1)<f(2).
- Let f is increasing in the interval [1;k] and f(n)>f(k) for all
natural n>k. We will show, that f is increasing in the interval [1;k+1]
and that f(n)>f(k+1) for all natural n>k+1.
Because f is increasing in the interval [1;k] and f(1)>=1, it is
easy to see that f(k)>=k. Let m is such point that f(m)>f(k) and f(m)
is minimal:
k <= f(k) <= f(m-1) < f(m) .
But we also know that f(m)>f(f(m-1)). Then f(m-1) <=k (otherwise
f(m) would not be minimal), and thus m-1=k and m=k+1.
Then:
f(1) < f(2) < ... < f(k) < f(k+1) < f(n)
for all natural n>k+1 and we proved that f is increasing.
We have also shown that f(n)>=n for all natural n.
Consider that there exists such natural n that f(n)>n, and thus
f(n)>=n+1. Then f(f(n))>=f(n+1), because f is increasing
and we get the contradiction. That's why there exists no natural n for which
f(n)>n and f(n)=n for all natural n.