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such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
(
) are positive reals satisfying
be positive real numbers such that
inequality and 
for all real numbers and ,
and 
to get: 
Multiply by the denominator to get:
But 
, from which the desired inequality follows
(QM-AM),
(Consequence of AM-GM).
Prove that
simply 
(?). Which is equal to 
,as, 
.
Now to finish notice that 
.![\[\frac{xy+yz+zx}{xyz}\leq \frac{x^2+y^2+z^2}{xyz} = \frac{2(a^2+b^2+c^2)}{xyz}=\frac{1}{2xyz}.\]](http://latex.artofproblemsolving.com/e/4/5/e45703e33db02b679950ae36f1b44fa0c1c49299.png)
We now wish to prove that 
,
and its cyclic forms.
,
.![\[\sum_{cyc} \frac{1}{\sqrt{a^2+b^2}} \le \sum_{cyc} \frac{\sqrt{2}}{a+b}.\]](http://latex.artofproblemsolving.com/4/4/f/44f5c5dea38147d48b8a2d808d5803050c705f44.png)
Thus, it suffices to show that ![\[\sum_{cyc} \frac{1}{a+b} \le \frac{1}{(a+b)(b+c)(b+c)}.\]](http://latex.artofproblemsolving.com/3/1/5/31501a383f9cdf2b14d3022f081ac35f8d0e4b68.png)
Clearing the fractions and expanding, this is equivalent to 
.
,
,
and likewise for the others. The desired inequality then turns into 
We also have 
by AMQM again, so the desired conclusion follows.
