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.
be an even integer, and let 
denote the number of (positive) divisors of . does not divide 
.
such that 
for all 
a regular hexagon inscribed in a circle 
. Let 
be a point inside and 
its polar reflection wrt . The rays 
meet again at 
. Call 
the polygon formed by the vertices . Similarly construct the polygon 
using instead. and are congruent.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
with an area of 
, with 
. If 
, Hence find the value of 
.
such that for all 
we have:
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
Let 
be positive reals such that 
.Prove the inequality 
pieces of equal weights. Can Vova prevent Pasha's win?
be vertices of regular pentagon inscribed in a circle whose radius is 
and center is at 
. Find all possible values of 
such that 
,prove :
.
![$ \sqrt[3]{abc} \le \frac{a+b+c}{3}=1.$](http://latex.artofproblemsolving.com/f/c/d/fcdee3361c66fbf3696139b3eeaf0d4afeb653ce.png)
and 
are non-negatives such that 
Prove that:![\[ \frac {1}{2 + ab^2} + \frac {1}{2 + bc^2} + \frac {1}{2 + ca^2} \ge 1\]](http://latex.artofproblemsolving.com/7/0/5/70555ed6014bc67521c9ea1e8b3dedeb7e6d0f8b.png)
are nonnegative real numbers satisfying 
then
such that:
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
Prove that
Let 
Prove that
Prove that
such that 
,
.
![${ \sum\frac{1}{2ab^{2}+1}=\sum (1-\frac{2ab^{2}}{2ab^{2}+1})=\sum (1-\frac{2ab^{2}}{ab^{2}+ab^{2}+1})\geq\sum (1-\frac{2ab^{2}}{3\sqrt[3]{a^{2}b^{4}}})=\sum (1-\frac{2}{3}\sqrt[3]{ab^{2}}})\geq \sum (1-\frac{2}{3}\frac{a+b+b}{3})=\sum (1-\frac{2}{9}(a+2b))=3-\frac{2}{9}(3a+3b+3c)=3-2=1$](http://latex.artofproblemsolving.com/5/2/3/5235814ce188e7049e3cf3ce1e8f06a760bb0c38.png)
![$ \sum\frac{1}{2+ab^{2}}=\sum\frac{1}{2}(1-\frac{ab^2}{2+ab^{2}})=\sum\frac{1}{2}(1-\frac{ab^2}{1+1+ab^{2}})\geq\sum\frac{1}{2}(1-\frac{ab^2}{3\sqrt[3]{ab^{2}}})=\sum\frac{1}{2}(1-\frac{1}{3}\sqrt[3]{a^{2}b^4})\geq\sum\frac{1}{2}(1-\frac{1}{3}\frac{ab+ab+b^2}{3})=\frac{1}{2}(3-\frac{1}{9}(a^{2}+b^{2}+c^{2}+2ab+2bc+ac))=\frac{1}{2}(3-\frac{1}{9}(a+b+c)^{2})=1$](http://latex.artofproblemsolving.com/5/2/b/52bd7002205d7c1d3253d7b21d22ed565316e04b.png)
