AoPS Academy
be a function such that for every regular 
-gon 
we have 
. Prove that 
for all reals 
.
be a triangle (
). 
are the altitudes of the triangle. The bisector of 
intersects 
at 
. Let 
be a point such that 
and 
. Prove that 
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)
be an interior point inside a triangle such that 
. Is it always true that
?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -11.92, xmax = 15.4, ymin = -7.32, ymax = 6.98; /* image dimensions */
pen qqwuqq = rgb(0,0.39215686274509803,0);
draw(arc((0.16,1.14),0.6,-160.76606420920794,-120.97284232859343)--(0.16,1.14)--cycle, linewidth(0.6) + qqwuqq);
draw(arc((0.16,1.14),0.6,-120.97284232859343,-81.17962044797893)--(0.16,1.14)--cycle, linewidth(0.6) + qqwuqq);
draw(arc((4.6,3.9),0.6,217.78184207622843,231.23032461400564)--(4.6,3.9)--cycle, linewidth(0.6) + qqwuqq);
draw(arc((4.6,3.9),0.6,204.33335953845113,217.7818420762284)--(4.6,3.9)--cycle, linewidth(0.6) + qqwuqq);
/* draw figures */
draw((4.6,3.9)--(-7.12,-1.4), linewidth(0.6));
draw((-7.12,-1.4)--(0.52,-1.18), linewidth(0.6));
draw((0.52,-1.18)--(4.6,3.9), linewidth(0.6));
draw((0.52,-1.18)--(0.16,1.14), linewidth(0.6));
draw((0.16,1.14)--(-7.12,-1.4), linewidth(0.6));
draw((-2.048809148845591,-1.2539709440767053)--(4.6,3.9), linewidth(0.6));
draw((-1.2633228022644833,-1.2313522272903383)--(0.16,1.14), linewidth(0.6));
draw(arc((0.16,1.14),0.6,-160.76606420920794,-120.97284232859343), linewidth(0.6) + qqwuqq);
draw(arc((0.16,1.14),0.47,-160.76606420920794,-120.97284232859343), linewidth(0.6) + qqwuqq);
draw(arc((0.16,1.14),0.6,-120.97284232859343,-81.17962044797893), linewidth(0.6) + qqwuqq);
draw(arc((0.16,1.14),0.47,-120.97284232859343,-81.17962044797893), linewidth(0.6) + qqwuqq);
draw(arc((4.6,3.9),0.6,217.78184207622843,231.23032461400564), linewidth(0.6) + qqwuqq);
draw((4.225582586081088,3.5319828806233673)--(4.118606182104256,3.4268351322300434), linewidth(0.6) + qqwuqq);
draw(arc((4.6,3.9),0.6,204.33335953845113,217.7818420762284), linewidth(0.6) + qqwuqq);
draw((4.150259228996044,3.6291527388051197)--(4.0217618658520555,3.5517678070351537), linewidth(0.6) + qqwuqq);
/* dots and labels */
dot((4.6,3.9),dotstyle);
label("$A$", (4.68,4.1), NE * labelscalefactor);
dot((-7.12,-1.4),dotstyle);
label("$B$", (-7.42,-1.78), NE * labelscalefactor);
dot((0.52,-1.18),dotstyle);
label("$C$", (0.5,-1.58), NE * labelscalefactor);
dot((0.16,1.14),dotstyle);
label("$M$", (0.24,1.34), NE * labelscalefactor);
dot((-2.048809148845591,-1.2539709440767053),linewidth(4pt) + dotstyle);
label("$D$", (-2.28,-1.66), NE * labelscalefactor);
dot((-1.2633228022644833,-1.2313522272903383),linewidth(4pt) + dotstyle);
label("$E$", (-1.36,-1.66), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/0/4/b/04b5cd117ed8b54c9685f5ddd5939c45284084ef.png)
so we are done.
>
& 
be the intersection of & 
angle bisector,any point within ∆
and -apollonius circle works.Let 
be intersection of angle bisector with .Since is within apollonius circle and apollonius circle & apollonius circle don't intersect (as their radax is perp. bisector of ), lies within & 
.So, done.
& 
.
