[/quote]
,
such that 
divides 
for all positive integers 
and 
with 
.
,
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
be an acute triangle with 
,
be the intouch triangle with 
,
,
,
be the intersecttion of the perpendicular from 
to 
with 
,
.
and 
are concylic
,
,
.
and that 
.
and 
be the midpoints of sides 
,
and 
be the feet of the altitudes from 
and 
in 
,
be the intersection of line 
with the tangent line to 
.
be the intersection point, other than 
.
be the intersection of lines 
and 
.
with centre 
passes through 
respectively. Let 
,
is equidistant from 
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
,
,
,
,
satisfy 
and 
.
of 
such that 
.
and 
)
Prove that if lines 
are similar.
such that 
, where 
are non-negative reals such that 
.
.
be lengths of tangents from 
to the incircle of the quadrilateral which center is 
.
be the midpoints of 
resp. Prove that![\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]](http://latex.artofproblemsolving.com/e/7/0/e709acb8b17c0466a385ba8e7a89bb1bf16c72af.png)
is given. It turns out that for every positive integer number 
,
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
Y by Adventure10 and 1 other user
Let 
be real numbers. Prove that
![\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]](//latex.artofproblemsolving.com/0/9/f/09fee86fb6ae5286ff38443f1590d912c4831d80.png)
Greece
EDIT BY DARIJ:
Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
be real numbers. Prove that![\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]](http://latex.artofproblemsolving.com/0/9/f/09fee86fb6ae5286ff38443f1590d912c4831d80.png)
Greece
EDIT BY DARIJ:
Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
This post has been edited 1 time. Last edited by Valentin Vornicu, May 3, 2006, 3:18 PM
be real numbers. Prove that ![\[ \frac 1{a(1+b)} + \frac 1{b(1+c)} + \frac 1{c(1+a)} \geq \frac 3 { 1+abc}. \]](http://latex.artofproblemsolving.com/e/c/e/eceb19d255038b777dc8cda4b96154efee7c590b.png)
.
where x is the best real .
where y is the best contsnt .![$[\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}][a(b+1)+b(c+1)+c(a+1)]\geq xy\geq 9$](http://latex.artofproblemsolving.com/3/2/f/32f31749d01431fa00a53ce2ac07882eac017b35.png)
and so 
and so
,
now could you please see my solution and tellme if it is correct ? i have a kind of battle with someone and i have to know from people like you if my solution is correct 
and 
.
and 
does not imply that 
.
are the "best" constants, you have 
,
for a certain constant 
?
would be the best constant, right? Which is equivalent to showing that there is always an equality case on the AM-GM for any 
so it only works for 
.
be positive real numbers . Prove that 
