[/quote]
,
be positive real numbers so that 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
,
,
cut three common chords concurrent at 
.
,
which are internally tangent to all 
.
that satises 
and 
for real 
can be written as 
where the greatest common divisor of positive integers 
,
,
is 
,
is not divisible by the square of any prime. Compute the value of 
.
and 
be positive real numbers such that 
.![\[
\frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)}
\geq \frac{3}{4}.
\]](http://latex.artofproblemsolving.com/5/6/5/56568d27a296f1c900fabbb1042549cd2e552e29.png)
,![\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1. \]](http://latex.artofproblemsolving.com/b/2/f/b2ff456632d4dd3b9ed93e4cba437bdd34d54cd8.png)
be a parallelogram of center 
.
,
and 
such that 
.
be given . On the extension of the segment 
beyond 
,
is chosen arbitrarily. Let the line through 
intersect the line 
at the point 
.
be the intersection of the lines 
and 
.
is tangent to the circumcircle of the triangle 
.
be any point. Let 
be projection point from 
.
cuts circumcircle 
at 
.
be antipode of 
are cyclic.
Y by
Can you sum
I coined a strategy Don Sirloin Bowel's algorithm because I found it and didn't see it anywhere else. Please tell me if this looks familiar because it can have absolutely scrumptious applications. (I hope that this stuff can help you sum that.) Starting with the "golden quadratic" or whatever the sigma it's called
. Multiplying by 
on both sides and resubstituting 
, we find 
. Continuing this process, we find 
, 
, and so forth. We claim that 
We already have our base case. 
. Multiplying by , 
. Hence, we have proved our claim by induction. Now, raising both sides to the 
th power,
So,
and
Are these identities useful? And can you use them to compute the sum?
I coined a strategy Don Sirloin Bowel's algorithm because I found it and didn't see it anywhere else. Please tell me if this looks familiar because it can have absolutely scrumptious applications. (I hope that this stuff can help you sum that.) Starting with the "golden quadratic" or whatever the sigma it's called
. Multiplying by 
on both sides and resubstituting 
, we find 
. Continuing this process, we find 
, 
, and so forth. We claim that 
We already have our base case. 
. Multiplying by , 
. Hence, we have proved our claim by induction. Now, raising both sides to the 
th power,
So,
and
Are these identities useful? And can you use them to compute the sum?This post has been edited 5 times. Last edited by BadAtCompetitionMath21420, Thursday at 5:36 PM
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