AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
is call 
if it satisfies the following properties:
and 
for all indices 
.
.
for which: Every sequence, there always exist two terms whose diffence is not less than 
. (where is given positive integer)
be a triangle with incentre 
and circumcentre 
. Let 
be the touchpoints of the incircle with 
, 
, 
respectively. Prove that 
is the Euler line of 
.
of integers that has the following property:
(possibly 
) then 
for every integer .
of nonzero integers such that the only admissible set containing both 
and is the set of all integers.
satisfies 
, and its incenter is . 
is the foot from to 
，and 
is a point on segment such that 
. Line 
meets 
at point 
, and the orthocenter of 
is 
.
is tangent to 
.
be a non-cyclic convex quadrilateral. The feet of perpendiculars from to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. Let the orthocenter of 
be . Prove that the common chord of circumcircles of 
and 
bisects 
.
and 
then is this uniformly convergence on 
comment on uniformly convergence on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
where in general it is should be uniformly convergence.
and trying to solve
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x^2 + y) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/c/7/c/c7cd7c9a4d4372afc23e4503bb1c05fc7d07e788.png)
be an acute-angled triangle with 
and let be the foot of the-angle bisector on . The reflections of lines and in line meet and at points and 
respectively. A line through meets and at 
and respectively such that and 
while lies strictly between 
and . Prove that the circumcircles of
and 
are tangent to each other.
of natural numbers such that
be a 
-dimensional inner product space of column vectors, where for 
and 
, the inner product of 
and 
is defined as ![\[\langle v, w \rangle = \sum_{i=1}^{10} v_i w_i.\]](http://latex.artofproblemsolving.com/9/f/9/9f9b7ff2fb19fee7e935c1b18dea3ec80547b9f9.png)
For 
, define a linear transformation 
on as follows:![\[ P_u : V \to V, \quad x \mapsto x - \frac{2\langle x, u \rangle u}{\langle u, u \rangle} \]](http://latex.artofproblemsolving.com/9/6/4/9646290e2b5e540b7f85d0d71e4d8992ccd8f542.png)
Given 
satisfying![\[ 0 < \langle v, w \rangle < \sqrt{\langle v, v \rangle \langle w, w \rangle} \]](http://latex.artofproblemsolving.com/d/8/0/d800fe48958cfee9ac01aa8bf11fe431ef45f8ef.png)
let 
. Then the dimension of the linear space formed by all linear transformations 
satisfying 
is 
that's manifestly true. Let 
be polynomial of degree 
, so
where 
is a monic real polynomial of degree 
and 
is a polynomial of degree with only real roots and real coefficients.
, where 
, so :
... (q.e.d)
![\[ P = b\prod^n_{i=1} (x-a_i)\]](http://latex.artofproblemsolving.com/4/f/f/4ff85858b0e206f5e31e0681d1cc6f1be3a8d9f7.png)
is not a root of 
and Let 
, then![\[ P' = P (c_1 + c_2 + \ldots + c_n), P'' = P (\sum_{i \neq j} 2c_i c_j)\]](http://latex.artofproblemsolving.com/0/3/1/0314153b422f93f1673a5b74fc59c59a3f2a82b3.png)
![\[ (n-1)(c_1 + c_2 + \ldots + c_n)^2 \ge 2n (\sum_{i \neq j}c_i c_j)\]](http://latex.artofproblemsolving.com/6/7/8/67818283bcdf49b5a6ba5660006db786af8a0c55.png)
![\[ \sum_{i \neq j} (c_i - c_j)^2 \ge 0\]](http://latex.artofproblemsolving.com/5/b/a/5ba21af2e92ced15e6f6e0817da00bf21ba68f5a.png)
are equal, which means all the 
are equal and 
be the zeroes (non-necessarily distinct) of . The given inequality holds trivially for 
. So we assume that 
in the latter part of the solution.![$P'(x) = P(x) [\sum_{i=1}^n\frac{1}{x-\alpha_i}]$](http://latex.artofproblemsolving.com/4/f/8/4f845e30f7bc73200410de9b56f19dd85763f6b7.png)
![$$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$$](http://latex.artofproblemsolving.com/7/0/7/70763c163e76fcfebb27909d534b3dcaf9420742.png)
can be re-written as ![$-n\left[\frac{P'(x)}{P(x)}\right]'\ge \left[\frac{P'(x)}{P(x)}\right]^2\iff 0\ge \frac{n(P''(x)P(x)-P'(x)^2)}{P(x)^2}+\frac{P'(x)^2}{P(x)^2}$](http://latex.artofproblemsolving.com/b/1/a/b1ae32c963595bc47ec0a1c85aca9f91e05a3ffe.png)
for all 
s are equal (equality case of Cauchy -Schewarz)
be the zeroes (non-necessarily distinct) of . The given inequality holds trivially for . So we assume that in the latter part of the solution. can be re-written as for all s are equal (equality case of Cauchy -Schewarz)
be the zeroes (non-necessarily distinct) of . The given inequality holds trivially for . So we assume that in the latter part of the solution. can be re-written as for all s are equal (equality case of Cauchy -Schewarz)
but you are excluding the roots.....this is the first problem so I can consider that 
but in that case if it's a constant polynomial then it has infinite roots...that is basically for all real ....then your statement is totally wrong....I don't think the 
idea is good for admitting all values of in real.....so if possible please post another solution....we are waiting for that
such that: 
.
.
