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 a fixed square and 
a variable point on segment 
. The square 
is constructed such that 
is on segment 
and 
is on segment 
. Let 
be the intersection point of lines 
and 
. Find the locus of as varies along segment .
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
with 
such that the equation 
has an integer solution 
.
such that
and 
Adit Aggarwal, India.
are smaller than the square root of , i.e., 
for all proper positive divisors 
of different from .
holds 
. does equality occur?
with![\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]](http://latex.artofproblemsolving.com/a/b/f/abf10d7e679f6f04db556424a0a0eda26f371fd4.png)
for all 
.
do there exist infinitely many positive integers such that 
divides 
?
such that has exactly 2000 prime divisors and divides 
?
positive real numbers such that 
. Prove that :
be a scalene acute triangle with incentre 
and circumcentre 
. Let 
cross 
at 
. On circle , let 
and 
be the mid-arc points of and 
, respectively. Let 
cross 
at 
and let 
cross 
at 
. Prove that the lines 
and 
are concurrent on the external bisector of 
.
of positive integers satisfying the following two conditions:
, if 
is in 
, then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
grid. Define a diagonal to be the six squares whose coordinates 
(
satisfy 
for some 
. Hence there are six diagonals.
(each exactly once) such that each row, each column, and each of the six diagonals has the same sum.
is a convex function on the interval ![$ [0, 1]. $](http://latex.artofproblemsolving.com/3/5/4/354f6da31e3233e3a6596f9e43cedb7ffa2099d1.png)
Note that 
for all ![$ x \in [0, 1] $](http://latex.artofproblemsolving.com/8/7/d/87da03d9bafefdb10c10753eccb2e719ecaf5fa6.png)
which implies the desired result.
for 
and suppose 
. Prove that for any positive integer 
,![\[ \left( a_1^k + \frac{1}{a_1^k} \right) \left( a_2^k + \frac{1}{a_2^k} \right) \dots \left( a_n^k + \frac{1}{a_n^k} \right) \ge \left( n^k + \frac{1}{n^k} \right)^n. \]](http://latex.artofproblemsolving.com/5/a/d/5ad4ea89463088b023d252d18d9f6eb411ec9f64.png)
is convex for 
for all 
with weights 
, we have 
, i.e.:
, again by Power Mean applied to positive reals 
with weights , we have 
i.e.:
So, we have 
. As a result, we need to prove that for any and 
that 
which hints towards taking the 
of both sides (credits to Eyed for telling me this smart way of Jensen-ing a bunch of products of a function). Doing that and dividing both sides by gets we need to prove 
so if is convex, we are done. Taking the derivative of gets us that for a constant , 
so the second derivative would be 
meaning since the bottom side is positive by trivial inequality and the top side is positive since 
if 
then the second derivative is positive meaning is convex, so as a result, which means that taking 
as the base and each side as an exponent gets 
for all positive real numbers 
and positive integer .
and 
, we have 
Proof: Note that if 
, we have 
. With this fact, the inequality follows by expansion since 
and 
.
while the expression 
monotonically decreases. At the state where all , the desired inequality achieves equality, so we're done.
then 
which is true because 
.
by AM-GM.
to be convex, so do the cancer operation on ur head (not writing it for the sake of my sanity)![$$\frac{\text{d}^2}{\text{dx}^2} \ln \left(x+\frac{1}{x} \right)=\frac{-x^4+4x^2+1}{x^2(x^2+1)^2} \implies \; \text{that thing is convex on} \; [0,1]$$](http://latex.artofproblemsolving.com/7/1/1/7119fad47f81d88b0bc38702c2a68bcfdcfd5930.png)
Since "that thing" is convex on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
we can use jensen to get
And by elevating by we are done, equality if 
thus we did it 
Notice that the second derivative of 
is 
Thus Jensen works.
both sides. Hence, it is enough to show that 
Let 
We will now prove that 
is convex. Indeed, we have 
and hence 
for every positive integer and 
Hence, we get 
However, 
and 
since 
, so we are done.
for every positive integer and . Hence, is convex and we can apply Jensen. We have 
so we are done. Equality occurs when 
But, note that the function 
is convex for 
, therefore by Jensen's we are done. QED
.
, s.t :
,
,![\[\prod_{k=1}^{n}(x_i+\dfrac{1}{x_i})\ge(\dfrac{k}{n}+\dfrac{n}{k})^n\]](http://latex.artofproblemsolving.com/e/6/3/e6397a21cf80d3b241f5f6f19b4c0f4e946d2785.png)
is equal to :
which is positive for all 
.