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with 
, a tangent to the circumcircle of at point 
intersects the extension of 
at point 
. 
is the midpoint of 
, and 
intersects the circumcircle of at 
. Prove that 
.
be positive real numbers such that 
. Prove that![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
, we have 
.
and 
are bisectors of triangle ABC. Prove that the segment 
intersects the inscribed circle of triangle 
.
on the segment 
we have 
where 
is the distance from point 
to line 
.. 
in 
we have 
then and 
lie in different sides of .
be the tangency point of the incircle with and 
be diameter in it. Then according to the remark above it is enough to check that . Denote 
etc., 
and 
. Let the tangent at 
intersect 
at 
, respectively. Then the triangles and 
are similar with coefficient 
. Thus we need to prove![\[2r>\frac{p-c}{c}\left(\frac{p-b}{c}h_b+\frac{p-a}{c}h_a\right)
\qquad \iff \qquad
\frac{c}{p-c}>\frac{p-b}{b}+\frac{p-a}{a}.\]](http://latex.artofproblemsolving.com/0/6/6/06611f316f62af557c08672c668f6051c3534914.png)
After Ravi's substitution 
it becomes![\[\frac{x+y}{z}>\frac{y}{x+z} + \frac{x}{y+z}
\qquad \iff \qquad
x\left(\frac{1}{z} - \frac{1}{y+z}\right) + y\left(\frac{1}{z}-\frac{1}{x+z}\right)>0\]](http://latex.artofproblemsolving.com/8/1/f/81fb7b968387454da8df250b7d24679fe50e6fd9.png)
which is obvious.
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
be the incenter and 
the inradius. Note that 
and 
.
,
and 
both lie outside the incircle.