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satisfying the condition![\[
2f(n)^2 = n^2 + g(n)^2 \quad \text{for all } n \in \mathbb{N}^*.
\]](http://latex.artofproblemsolving.com/6/a/1/6a112ea3fe44273fdd4dc445e9bd7387d365925f.png)
Prove that if 
for all 
, then 
has infinitely many fixed points.
be a diameter of a circle 
, and let 
be the tangent at 
. Furthermore, let 
be a fixed, positive real, and consider all pairs of points 
and 
lying on , on opposite sides of , such that 
. The lines 
and 
intersect at points 
and 
, respectively. Show that all the lines 
pass through a common point.
and 
must be fixed for it to happen, because in the proof I don't see where we use it
corresponding to 
, 
,
if and only if 
(Hence 
, and we will implicitly assume 
in the following.). Setting 
it is equivalent to 
for some 
. (Remember we have fixed 
) Since 
we see 
. Since there exists 
such that 
and obviously 
for sufficiently large 
, we conclude that it must have at least one positive root. Let α be the largest positive root, then it is obvious that 
p(α) =
where 
α. That is: The biggest value can attain is α, so 
α. This can be attained by the previous proof, and it does so only when two of 
are equal (because 
). Let β be the smallest positive root of 
.If 
when 
[0;β) then is at least β and 
β,with equality only if two of 
are equal, since . If when 
[0;β) then 
with equality when one of are zero,since 
.
so we can only use it when proving inequalities for all! non-negative .If you still don't believe that read the "proof" of following inequality using ABC theorem: be non-negative reals such that 
Prove that 
:
or 
.In these cases we have ![$0=abc(a-b)^2(b-c)^2(c-a)^2=24==>1=0==>abc=abc-[abc]-100<4$](http://latex.artofproblemsolving.com/1/5/6/1568d6d8d89e4475252db19b7b05a3c1051857bb.png)
.Done!!
method!
?
s.t 
