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.
decide whether there exist pairwise distinct positive integers 
such that for every 
, 
divides 
.
+1 w
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
, consider the 
fractions 
The product of these fractions equals 
, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
be a function satisfying
for all integers 
and 
. Show that there exist positive integers 
and such that 
for all integers .
of the triangle 
which doesn't coincide with 
or 
intersectes the altitudes from 
and 
at 
and 
respectively. Let 
be the reflection of in and 
be the reflection of in 
Prove that the circles 
and 
are tangent.
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
such that for any positive integer 
there exists a positive integer such that the last digits of 
are equal to 
such that for any positive real numbers 
, which are the sides of a triangle, the following inequality holds:![\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]](http://latex.artofproblemsolving.com/9/1/c/91c8a4af2bd2f3107c8921d2e06f8a09dedd0164.png)
be a non constant polynomial with real coefficients. Prove that the system of simultaneous equations —
has finitely many solutions .
such that 
.
then using Mean Value Theorem, there exist an such that 
(Maybe it can't be prove: you can construct a function with 
and 
for 
and max 
). I hope the statement was 
because, in that case, the same as we have done.
case, suppose 
and assume 
on the whole interval 
. Then by the mean value theorem, there is some such that 
so 
. That by itself isn't a contradiction, but we can do something similar to get a contradiction. The intuitive idea is that since we're right on the "edge" of a contradiction, we have no room to move the graph of 
. If 
and , that forces 
on the interval ![$[0,1/2]$](http://latex.artofproblemsolving.com/8/2/4/8240ac9beaadf3f2654c3ba09a3987a7305539c5.png)
. Moving the graph up or down at any point makes the slope larger than 
somewhere. Similarly, 
on the interval ![$[1/2,1]$](http://latex.artofproblemsolving.com/5/5/8/5581510e950d8ceb9bcbcfd0dc988d9c4538c684.png)
. for every ![$x\in [0,1/2]$](http://latex.artofproblemsolving.com/0/f/d/0fd16eeab1cb0c53e329c87cea809db55ea5f207.png)
. On any interval ![$[0,a]$](http://latex.artofproblemsolving.com/6/8/b/68bacdc4b96052586744d16169ccd8c3233ed09f.png)
with 
, we have 
for some 
, depending on 
. Since we're assuming 
is always bounded by , this means that 
so 
for all ![$a\in[0,1/2]$](http://latex.artofproblemsolving.com/e/0/c/e0c5302f45677b07a169c44aa1085cf1dace8463.png)
.
, then we could use the mean value theorem on the interval ![$[a,1/2]$](http://latex.artofproblemsolving.com/f/7/c/f7c796537cc023b8e3d0dd550cf1c3bdfc0278d4.png)
to get 
But the left side is larger than , and the right side is at most , a contradiction. on the interval , so 
However, this function is not differentiable at 
, so we are done.
:
Assume towards a contradiction that 
.
and 
:![\[\usepackage{mathtools}
\tfrac {1-f(x)}{\frac 12-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\tfrac 12\right):f(x)\ \ge\ 2x\tag{1}\]](http://latex.artofproblemsolving.com/4/6/d/46d07da0b23e15de91ca1912093a406842150f21.png)
LMVT on and 
:![\[\usepackage{mathtools}
\tfrac {f(x)-1}{x-\frac 12}\ge -2\xRightarrow{\hspace{40pt}}\left(\forall \tfrac 12<x< 1\right):f(x)\ \ge\ 2-2x\tag{2}\]](http://latex.artofproblemsolving.com/9/2/c/92c82c921fad2971d842ec2dba87a5e4725c3e97.png)
Differentiability at 
implies 
is not possible.
and 
, reflecting about the line if necessary, we may assume 
.
and 
:![\[\usepackage{mathtools}
\tfrac {2\alpha+\eta-f(x)}{\alpha-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\alpha\right):f(x)\ \ge\ 2x+\eta\tag{3}\]](http://latex.artofproblemsolving.com/8/3/5/83584e1961403c7cf5c6891281a4cbb97346d193.png)
continuity at 
:![\[(\forall \epsilon>0)(\exists\delta>0):0<x<\delta\xRightarrow{\hspace{40pt}}\big|f(x)\big|<\epsilon\tag{4} \]](http://latex.artofproblemsolving.com/9/d/a/9da78481512b68b5c3e9ed771a83552d9878e137.png)
contradicts 
for 
.
is differentiable on . For 
and 
you can hopefully do!! put the value of 
in the two equations,So clearly visible that not differentiable on ....
and 
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and 
and 
.
such that: 
.
be a function which is continuous on ![\( [0,1] \)](http://latex.artofproblemsolving.com/0/1/0/01026be294289be8f090c7b2586dde64a0d5b479.png)
and differentiable on 
,
.
such that 
.
such that 
