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.
are positive real numbers such that 
, prove that ![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](http://latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
, denote the sum of digits of in its decimal representation by 
. Find all polynomials 
with integer coefficients such that for any positive integer 
, the integer 
is positive and 
of prime numbers which 
and
is an integer.
children sit in a circle and play the game described below. The teacher goes around the children clockwisely and hands out candies to them according to the following regulations: Select a child, give him a candy; and give the child next to the first child a candy too; then skip over one child and give next child a candy; then skip over two children; give the next child a candy; then skip over three children; give the next child a candy;... for which the teacher can ensure that every child get at least one candy eventually (maybe after many circles).
, and 
be positive real numbers, such that 
. Prove the inequality![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
such that![\[
x+f(xy)=f(1+f(y))x
\]](http://latex.artofproblemsolving.com/4/3/3/433f95836dad1224754d480aff8432ef033a37cb.png)
for all 
.
such that![\[ \left\lfloor\frac{n+1}{x}\right\rfloor = n - \left\lfloor \frac{n}{x} \right\rfloor + \left \lfloor \frac{\left \lfloor \frac{n}{x} \right\rfloor}{x}\right \rfloor - \left \lfloor \frac{\left \lfloor \frac{\left\lfloor \frac{n}{x} \right\rfloor}{x} \right\rfloor}{x}\right \rfloor + \cdots \]](http://latex.artofproblemsolving.com/3/1/9/319c9a2a80c4da5961cd5389af0606fa429ca8e2.png)
for any positive integer .
such that for all positive real numbers 
and 
:
that satisfy the equation![\[
f(x) + 3f(-x) + f\left( \frac{1}{x} \right) = x, \quad \text{for all } x \in \mathbb{R}^*.
\]](http://latex.artofproblemsolving.com/6/a/2/6a2e35c42644036a62451b5e1794b36ffa728591.png)
be the set of positive integers. Determine all functions 
such that 
is divisible by 
for all positive integers 
.
from positive integers to themselves st:
holds for any ?
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
to deduce all of them are not divisible by 
denote the k-th symmetric sum and 
that 
and by newton sums we have 
and we consider the polynomial
which in 
is 
and the roots are 
so we get 
but 
and we are done
we get 
so as 
we have 
as desired.
, consider the polynomial 
, clearly 
and ![$P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/0/5/a05997bd7ba17001c62bc17ae35c76dd78e24789.png)
. And therefore we can let 
where all 
's are integers, so by taking newton sums modulo we have 
which gives that 
therefore by putting 
in this we get that 
for any 
which proves that all 's are congrent modulo and therefore we are done 
.
. Suppose 
of the 's are nonzero and WLOG they are 
. Let 
be a generator of 
. Let 
where 
is an integer for 
. Then the condition becomes 
for all 
. It follows that the elements of are roots of ![$P(x):=x^{\alpha_1}+\cdots+x^{\alpha_r}\in\mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/8/b/0/8b0a5b5937b0aff17391fefd8df1603fb6c521a8.png)
. Thus is divisible by![\[
\prod_{a\in\mathbb{F}_p^\times}(x-a)=x^{p-1}-1.
\]](http://latex.artofproblemsolving.com/c/9/d/c9d1ccf7462e298113392f4798239e5f502ae2f4.png)
Since 
if 
is nonzero, 
and thus either 
in which case we are done, or 
and all the 
's are equal, as desired. 
Prove that
be a prime number and 
be integers for which 
is divisible by 
.
is divisible by 
)
denote the 
,
.
(interpreted in ![$\mathbb F_p[t]$](http://latex.artofproblemsolving.com/d/1/f/d1fa992d9fe693e162a54b2121d646af3dd9f0e2.png)
)
,
.
.
But polynomials in 
modulo