AoPS Academy
be the incenter of triangle 
, and 
be the perpendicular bisector of 
. Suppose that 
is on the circumcircle of triangle , and line 
and intersect at point 
. Point 
is on such that 
.Suppose that line 
and the midsegment of that is parallel to 
intersect at 
. Show that 
is defined recursively by ![\[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]](http://latex.artofproblemsolving.com/f/4/b/f4b8f95e49de7274cccd8ee498e0b496f31ddfea.png)
Show that 
for all 
.
be positive real numbers. Prove that:![$$\sqrt[3]{a^3+b^3}+\sqrt[3]{b^3+c^3}+\sqrt[3]{c^3+a^3}\geq \sqrt[3]{2}(a+b+c)$$](http://latex.artofproblemsolving.com/2/6/e/26ebcdae35153bee73abbcfaf4a767d3a32b8912.png)
-plane and a tiling of this plane with white and red unit squares in a checkerboard pattern (ensure the vertices of the squares are at lattice points). For each square, draw a regular tetrahedron with the property that each vertex of the square coincides with the midpoint of one of the edges of the tetrahedron. Notice for every square there are exactly two distinct possible ways the tetrahedron could be drawn. On the white squares ensure that the upper edge of the tetrahedron is parallel to the 
-axis and for the red squares ensure that the upper edge of the tetrahedron is parallel to the 
-axis.
rotation about the or axis it is clear that the tetrahedrons fit together seamlessly and their vertices coincide with the midpoints of the edges of adjacent tetrahedron.![[asy]
size(200);
import three;
currentprojection = perspective(3,5,7);
draw(surface((-4,-4,0)--(4,-4,0)--(4,4,0)--(-4,4,0)--cycle), gray+opacity(0.2));
draw((-4,0,0)--(4,0,0), black); // x-axis
draw((0,-4,0)--(0,4,0), black); // y-axis
triple[] tetra1 = {(1,-1,1), (1,3,1), (-1,1,-1), (3,1,-1)};
triple[] tetra2 = {(-1,-3,1), (-1,1,1), (-3,-1,-1), (1,-1,-1)};
triple[] tetra3 = {(-3,1,1), (1,1,1), (-1,-1,-1), (-1,3,-1)};
triple[] tetra4 = {(-1,-1,1), (3,-1,1), (1,-3,-1), (1,1,-1)};
void drawTetra(triple[] T, pen color) {
draw(surface(T[0]--T[1]--T[2]--cycle), color+opacity(0.4));
draw(surface(T[0]--T[1]--T[3]--cycle), color+opacity(0.4));
draw(T[0]--T[2]--T[1]--T[3]--cycle, black+linewidth(0.8));
draw(T[0]--T[1], black+linewidth(0.8));}
drawTetra(tetra2, red);
drawTetra(tetra4, grey);
drawTetra(tetra1, red);
drawTetra(tetra3, grey);
draw(surface((-1,-1,1)--(-1,1,1)--(0,0,0)--cycle), red+opacity(0.4));
draw(surface((-1,-1,1)--(-1,1,1)--(-2,0,0)--cycle), red+opacity(0.4));
draw((-1,-1,1)--(-1,1,1), linewidth(0.8));
draw((2,2,0)--(2,-2,0)--(-2,-2,0)--(-2,2,0)--cycle);
draw((2,0,0)--(-2,0,0));
draw((0,2,0)--(0,-2,0));
draw((0,0,0)--(-1,1,1), linewidth(0.8));
draw((0,0,0)--(-1,-1,1), linewidth(0.8));
draw((-2,0,0)--(-1,1,1), linewidth(0.8));
draw((0,0,-4)--(0,0,4), black, Arrow3);
label("$z$", (0,0,4.5), N);
label("$x$", (4.5,0,0), E);
label("$y$", (0,4.5,0), N);
[/asy]](http://latex.artofproblemsolving.com/8/4/1/84127fa541e6ee5089fdd53383a2ee3fadc4cebe.png)
-plane do this for all planes 
where 
is a non-zero integer (we are "stacking" the sheets of tetrahedron). It is not hard to see that after drawing such tetrahedron, we are left with a collection of octahedron in space. We now go through a separate procedure for partitioning each of these octahedron into tetrahedron.-plane we can simply divide it into four tetrahedron. Otherwise let the octahedron touch the -plane at . Call the octahedron 
. Then let 
be the result of after applying the homothety centered at with factor 
. Then 
can be tiled with a finite number of tetrahedron (specifically 
). Then repeat this process with and so on.-plane intersects each of the original tetrahedron at exactly 
points and non of the tetrahedron formed by portioning the octahedron touch the -plane, we are finished.
![[asy]
import geometry;
size(2cm);
pair A=dir(90); pair B=dir(210); pair C=dir(330);
pair A1=.5(C+B); pair B1=.5(A+C); pair C1=.5(A+B);
pair A2=.5(B1+C1); pair B2=.5(A1+C1); pair C2=.5(A1+B1);
pair A3=.5(B2+C2); pair B3=.5(A2+C2); pair C3=.5(A2+B2);
pair A4=.5(B3+C3); pair B4=.5(A3+C3); pair C4=.5(A3+B3);
filldraw(A--B1--C1--cycle, red+opacity(.5));
filldraw(B--A1--C1--cycle, red+opacity(.5));
filldraw(C--B1--A1--cycle, red+opacity(.5));
filldraw(A1--B2--C2--cycle, red+opacity(.5));
filldraw(B1--A2--C2--cycle, red+opacity(.5));
filldraw(C1--B2--A2--cycle, red+opacity(.5));
filldraw(A2--B3--C3--cycle, red+opacity(.5));
filldraw(B2--A3--C3--cycle, red+opacity(.5));
filldraw(C2--B3--A3--cycle, red+opacity(.5));
filldraw(A3--B4--C4--cycle, red+opacity(.5));
filldraw(B3--A4--C4--cycle, red+opacity(.5));
filldraw(C3--B4--A4--cycle, red+opacity(.5));
dot((0,0));
[/asy]](http://latex.artofproblemsolving.com/9/1/8/9181484fa01d455685df228ad52b299a31e7d185.png)
and 
such that
