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 real numbers such that ![\[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \]](http://latex.artofproblemsolving.com/8/f/5/8f5d49c9b44e3db3cd85fea7b6521e1dc02d4269.png)
Prove that ![\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]](http://latex.artofproblemsolving.com/c/0/4/c0438d6eba686d0a11fd4decb4afa9a14aa50bac.png)
are non-negative integers.
satisfying the following equation:
.
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)
children, no two of the same height, lined up. Let us call a pair of different children 
good if between them there is no child whose height is greater than the height of one of 
and 
, but less than the height of the other. What is the greatest number of good pairs that could be formed? (Here, and 
are considered the same pair.)
satisfy 
prove in equality.
![\[\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),\]](http://latex.artofproblemsolving.com/0/4/7/047adb616a6be9f9ed52f2385fa592fb5464e420.png)
and 
.
and 
be elements of the set 
such that 
, 
is a divisor of 
, and 
is a divisor of 
. Determine the largest value that can take.
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
have the property that 
is a perfect square for infinitely many positive integers 
?
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
. Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
that is not isosceles at 
, let 
be its incircle, which is tangent to 
at 
, respectively. The lines 
and 
intersect the line passing through and parallel to 
at 
and 
, respectively. The lines passing through 
and perpendicular to 
intersect 
and 
at 
and 
, respectively.
are collinear and that 
are concurrent.
positive integer numbers such that the least common multiple (LCM) of all these numbers is not a perfect square. Mykhailo consecutively hides one of these numbers and writes down the LCM of the remaining 
numbers that are not hidden. What is the maximum number of the written numbers that can be perfect squares?
be a family of functions from 
. It is known that for all 
, there exists 
such that for all 
, the following equation holds:![\[
y^2 \cdot f\left(\frac{g(x)}{y}\right) = h(xy)
\]](http://latex.artofproblemsolving.com/f/7/1/f7196164e0639695529b06788ea7cb1e50eddbe7.png)
and all 
, the following identity is satisfied:![\[
f\left(\frac{x}{f(x)}\right) = 1.
\]](http://latex.artofproblemsolving.com/1/5/b/15b363bf1754a4edd72423c3d0ea631173897beb.png)
. By hypothesis, there exists a function 
such that![\[
y^2\cdot f\left(\frac{f(x)}{y}\right) = g(xy), \quad \forall x,y\in\mathbb{R}^+
\]](http://latex.artofproblemsolving.com/4/d/e/4de11b6a7f4cdaae20e844da9ed8dd948db338bd.png)
Substituting 
gives![\[
f^2(x) f(1) = g(x f(x)), \quad \forall x \in \mathbb{R}^+
\]](http://latex.artofproblemsolving.com/c/a/3/ca3e23d972be3e9aacb9e6e4e757ee23529306c2.png)
Similarly, another function 
satisfies![\[
y^2 g\left(\frac{f(x)}{y}\right) = h(xy), \quad \forall x,y\in\mathbb{R}^+
\]](http://latex.artofproblemsolving.com/6/3/2/6329b9ac5863c54ef921dc081c23d38585bfe534.png)
Setting 
, we obtain![\[
g(x f(x)) = h(1) x^2, \quad \forall x \in \mathbb{R}^+
\]](http://latex.artofproblemsolving.com/8/b/0/8b015c73d1dd88542a49c1cf7a46a29f5156b7e1.png)
From the previous equation,![\[
f^2(x) f(1) = h(1) x^2
\]](http://latex.artofproblemsolving.com/6/e/a/6eae3c8779d1e0b12e1b4d99a91c925fd4454a88.png)
Rearranging,![\[
f^2(x) = \frac{h(1)}{f(1)} x^2
\]](http://latex.artofproblemsolving.com/f/4/a/f4af16541a2c512c7a9d83f077afac0142939394.png)
which implies![\[
f(x) = c x, \quad \forall x\in\mathbb{R}^+
\]](http://latex.artofproblemsolving.com/8/1/1/8119b15f1885facd586b4bbf47a3b0aea2d51588.png)
for some constant 
. Finally,![\[
f\left(\frac{x}{f(x)}\right) = f\left(\frac{1}{c}\right) = 1
\]](http://latex.artofproblemsolving.com/f/1/a/f1ab747b7e7d797d498ed99f4babf933b23dc6ed.png)
as required.
we get an 
and a equation, in which we set 
to get that 
, now by creating an 
for the queried 
we get by letting 
that 
which now means that 
and thus 
for all positive reals 
, and this is the shape of any random function in 
, so the condition now holds trivially as 
, and thus we are done 
and let 
and 
be tangency points of the incircle of 
,
and 
respectively. Let the circle with centre 
passing through 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
