)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
and 
.
passes through the midpoint of 
.
& 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
Y by Adventure10, Mango247
Let 
be a triangle with 
and the points ![$D \in [AC], E\in [BC]$](//latex.artofproblemsolving.com/7/3/9/739ae483b177297721e9c87a4604576b92937d00.png)
. Inside the triangle we construct the semicircles 
of diameters ![$[AC], [BC], [CD], [CE]$](//latex.artofproblemsolving.com/2/a/a/2aaabd6982ac6df138ee457796c4cb9161c5abf2.png)
and let 
. Show that points 
are concyclic.
be a triangle with 
and the points ![$D \in [AC], E\in [BC]$](http://latex.artofproblemsolving.com/7/3/9/739ae483b177297721e9c87a4604576b92937d00.png)
. Inside the triangle we construct the semicircles 
of diameters ![$[AC], [BC], [CD], [CE]$](http://latex.artofproblemsolving.com/2/a/a/2aaabd6982ac6df138ee457796c4cb9161c5abf2.png)
and let 
. Show that points 
are concyclic.This post has been edited 1 time. Last edited by parmenides51, Dec 24, 2018, 12:37 AM
Reason: added ''{C, M} =C_3 \cap C_4 '', after the first comment
Reason: added ''{C, M} =C_3 \cap C_4 '', after the first comment
at the end, thanks for noting the missing info
Consider a triangle 
.
and 
be the circles with diameter 
and 
Thus, 
is the foot of the height from 
to 
.
Let 
be the foot of the height from 
is the circumcircle of 
too, 
is the circumcircle of 
So 
is the radical axis of 
we get that: 
,
,
,
and 
Take 
,
Note that: 
Too we have that: 
Besides that, the following quadrilaterals are cyclic, because of the perpendicularism said over: 
and 
,
now calculating 
Therefore, 
so 
is cyclic 
