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.
denote the set of positive integers
such that for all 
it holds that 
holds for all real numbers 
.
+1 w
be positive real numbers such that 
Prove that:
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
, with initial values 
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
and write the numbers 
down on a board. A move consists in erasing any two numbers 
and 
, then writing down the numbers 
and 
on the board, and then removing repetitions (e.g., if the board contained the numbers 
, then one could choose the numbers 
and 
, obtaining the board with numbers 
). For all integers , determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves. and a polynomial 
with real coefficients of degree 
(
) such that has no real roots. Prove that the polynomial![\[
Q(x) \;=\; P(x) \;+\; a\,P'(x) \;+\; a^2\,P''(x) \;+\; \dots \;+\; a^n\,P^{(n)}(x)
\]](http://latex.artofproblemsolving.com/6/5/e/65e5e4eefcadd5d0019b4929b3472cdf203c13aa.png)
has no real roots.
and 
, with 
, a set 
the pair 
if for any elements 
we have 
. Let 
be the greatest possible number of elements of a set that escapes the pair . Find the maximum and minimum values of 
.
with incircle 
, let 
be the incenter and 
be the point where touches 
. Let 
be the point on 
with 
and 
be the orthocenter of 
, so that 
on 
also satisfies 
. Prove that 
, the reflection of over the midpoint of 
, lies on .
satisfying 
the following inequality holds:
of the triangle consider a mobile point 
. Let 
the orthogonal projection of 
on 
. If the mobile points 
and 
are such that 
is a paralellogram, find the locus of point 
when goes through 
.
be a positive integer. There are 
real coefficient polynomials 
which is not all the same, and their leading coefficients are positive. Prove that
and find when the equality holds.
be cyclic homogeneous polynomial with variables, 
, 
and 
as above. Prove that:
be a positive integer. There are real coefficient polynomials which is not all the same, and their leading coefficients are positive. Prove thatand find when the equality holds.. In case 
are not of the same degree, the degree of is 
and we are done, so assume 
. Denote by 
the senior coefficient of . The coefficient at 
equals
(by a well known inequality). Moreover, the equality is attained only when all 
are equal. Further, we assume 
. Let 
We have
and
Now, putting 
in (1) and (2) yields
Let 
and 
be the coefficient at 
of 
. Let's put 
in (1) and (2). The coefficient at 
of the polynomial equals
which can be written as![$$\frac{n}{2}\left((n-1)\sum_{i=1}^n q_i^2 - \sum_{i_1,i_2\in [1..n]}q_{i_1}q_{i_2} \right)$$](http://latex.artofproblemsolving.com/0/d/d/0ddd0eccc01a2dcd6cbe041b1d0b43242536c8f0.png)
Note that the sum in the big brackets is not negative, and equals 
only when 
are equal. Let 
be the greatest integer such that the coefficients at 
of 
are not all equal. In the same way, we obtain that the coefficient at 
of is positive. Therefore
The equality is attained when 
, that is, when 
and 
are not all equal.
be cyclic homogeneous polynomial with variables, , and as above. Prove that:
and 
and put 
be a positive integer. There are real coefficient polynomials which is not all the same, and their leading coefficients are positive. Prove thatand find when the equality holds.![$P_1, \dots, P_n \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/5/d/05dea8985af8f0f9e54219346a3427b00fc365a8.png)
, which is not all the same, with positive leading coefficient such that![\[ \deg(P_1^n + \dots + P_n^n - nP_1 \cdots P_n) < (n - 2) \max_{1 \le i \le n}(\deg P_i) \]](http://latex.artofproblemsolving.com/d/2/3/d235bcdf8c9f4226776f80025e01b4a9479dacbf.png)
WLOG 
.
, then as all has positive leading coefficients, we have![\[ \deg(P_1^n + \dots + P_n^n) \ge n\deg(P_1) > \deg(P_1\cdots P_n) \implies \deg(P_1^n + \dots + P_n^n - nP_1 \cdots P_n) = \deg(P_1^n + \dots + P_n^n) \ge n \deg(P_1) \]](http://latex.artofproblemsolving.com/d/a/c/dac093ac5ceb39bd444609dd67e4e270c663dc73.png)
which is a contradiction. Therefore, we must have 
. Write 
, where 
. The condition above implies that ![$[x^{\ell}] (P_1^n + \dots + P_n^n - nP_1 \cdots P_n) = 0$](http://latex.artofproblemsolving.com/1/5/8/15857388bff8df57d3f41525089768389f78e633.png)
for all 
. \\
&= [x^{mn}] \left( \sum_{i = 1}^n (a_i x^m + Q_i)^n - n \prod_{i = 1}^n (a_i x^m + Q_i) \right) \\
&= \sum_{i = 1}^n a_i^n - n \prod_{i = 1}^n a_i \stackrel{\text{AM-GM}}{\ge} 0
\end{align*}](http://latex.artofproblemsolving.com/c/f/8/cf87ea5f80f2d25ef7ba55dda08882b2e19fd4f1.png)
and thus, we must have 
, and 
since 
are not all the same. Now, let 
be the largest positive integer such that the coefficients of in 
is not all the same, which must exist by the hypothesis of the problem.
for each 
, where ![$R \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/d/a/4da527ded09c721d368e1335555f0d97d0b81791.png)
, 
and 
. Thus, 
are not all the same by our choice of 
. First of all, note that for any 
,![\begin{align*}
\deg ( (x^rR)^{n - \ell} (r_ix^r + S_i)^\ell ) &= (n - \ell) \deg (x^rR) + \ell \deg (r_i x^r + S_i) \le (n - \ell)m + \ell r < m(n - 2) + 2r \\
\deg \left( \sum_{S \subseteq [n], |S| = \ell} (x^r R)^{n - \ell} \prod_{i \in S} (r_i x^r + S_i) \right) &\le \max_{1 \le i \le n} (n - \ell) \deg (x^rR) + \ell \deg(r_i x^r + S_i) \le (n - \ell)m + \ell r < m(n - 2) + 2r
\end{align*}](http://latex.artofproblemsolving.com/a/6/5/a657c25f03b66cb9b235249cf742b5a20e69debe.png)
Therefore, we conclude that![\begin{align*}
0 &= [x^{m(n - 2) + 2r}] (P_1^n + \dots + P_n^n - nP_1 \cdots P_n) \\
&= [x^{m(n - 2) + 2r}] \left( \sum_{i = 1}^n (x^rR + (r_ix^r + S_i))^n - n \prod_{i = 1}^n (x^rR + (r_ix^r + S_i)) \right) \\
&= [x^{m(n - 2) + 2r}] \left( \binom{n}{2} (x^rR)^{n - 2} (r_i x^r + S_i)^2 - n (x^r R)^{n - 2} \sum_{1 \le i < j \le n} (r_ix^r + S_i)(r_j x^r + S_j) \right) \\
&= \binom{n}{2} a^{n - 2} r_i^{2} - n a^{n - 2} \sum_{1 \le i < j \le n} r_i r_j \\
&= \frac{n}{2} \cdot a^{n - 2} \cdot \left( \sum_{1 \le i < j \le n} (r_i - r_j)^2 \right) \ge 0
\end{align*}](http://latex.artofproblemsolving.com/6/8/7/68742532b88c3bff81ea6aec58987979b4225747.png)
where last equality hold if and only if all s are equal, which is not the case. Therefore, this equality can't hold, and therefore our original assumption is false. Therefore, we must have![\[ \deg(P_1^n + \dots + P_n^n - nP_1 \cdots P_n) \ge (n - 2) \max_{1 \le i \le n}(\deg P_i) \]](http://latex.artofproblemsolving.com/5/e/e/5ee500419da61a1236420dc09325848f4234942c.png)
Equality holds when 
, i.e. we have 
for some 
not all equal.
, one can simply notice that for sufficiently large 
,![\begin{align*}
P_1^n+P_2^n+\cdots +P_n^n-nP_1P_2\cdots P_n &= \left(P_1^{n/2} - P_n^{n/2}\right)^2 \\&+ \left[\left(\sqrt{P_1P_n}\right)^{n} +\left(\sqrt{P_1P_n}\right)^{n} + \sum_{i=2}^{n-1} P_i^n - n \left(\sqrt{P_1P_n}\right)^2\prod_{i=2}^{n-1} P_i \right]
\\&\ge \left(P_1^{n/2} - P_n^{n/2}\right)^2
\\&\ge (P_1 - P_n)^2 P_n^{n-2},
\end{align*}](http://latex.artofproblemsolving.com/1/3/8/1385ef033276bc60d3c790ff9ae5203e4d9cc2c3.png)
where the last inequality follows from the mean value theorem. The desired result clearly follows.
be a positive integer, and 
be pairwise different real-coefficient polynomials with positive leading coefficients. Prove that
satisfying 
