AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
, centroid 
and orthocenter 
. Prove that 
is constant.
-
plane, given a parabola 
passing through 3 points 
and 
.
vary in the range of 
, draw the domain swept out by .
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
is defined by 
and 
Show that there exist infinitely many prime number that divide at least one number in this sequences
be an acute scalene triangle , 
be the orthogonal projection of 
on 
, 
and 
are the midpoints of 
and 
respectively. Let 
and 
are points on the minor arcs 
and 
of the circumcircle of triangle respectively such that 
. Show that the circumcircles of triangles 
and 
are tangent if and only if lies on 
.
be 
given points in the plane, and let 
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points , i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex 
a non-negative real number 
, for 
, such that 
. Bob then selects a sequence of distinct vertices 
such that 
and 
are connected by an edge for every 
. (Note that the length 
is not fixed and the first selected vertex always has to be 
.) Bob wins if![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)
otherwise, Alice wins. Depending on 
, determine the largest possible value of 
for which Bob has a winning strategy.
such that![\[
f(x^2 \cdot f(x) + f(y)) = f(f(x^3)) + y
\]](http://latex.artofproblemsolving.com/8/6/5/865b10bbc5ba7721078b1c60eaeede2abfeeb1a2.png)
. be an acute triangle and let be an interior point for it such that 
. Prove that
When does equality hold?
such that![\[
f(f(x)+y) = x+f(f(y))
\]](http://latex.artofproblemsolving.com/9/d/7/9d780faa5901646f7035e88400a9e3d8023f04ff.png)
for all 
.
, with 
, 
is the incenter, 
is the intersection of -excircle and . Point 
lies on the external angle bisector of 
such that and lieas on the same side of the line 
and 
. Point lies on such that 
. Circle 
is tangent to 
and at 
, circle 
is tangent to and at 
and both circles pass through the inside of triangle . if is the Midpoint od the arc , which does not contain , prove that lies on the radical axis of and . be a triangle with incircle 
, which touches sides 
and at points and , respectively. Choose points and on the line 
such that 
and 
. Let be the intersection of lines 
and 
. Define and 
as the intersections of 
and 
with lines 
and 
, respectively. Assume that 
is the intersection of 
and . Prove that 
.
points in a circle and number them arbitrarily with numbers from 
to . The chords cannot share common endpoints, also, the endpoints of these chords should be among the points.
pairwisely non-intersecting chords can be drawn for each of whom its endpoints differ in at most 
. cannot be improved.
satisfies that for any positive integer 
and 
Suppose there is 
non-negative reals 
such that for 
Determine the least possible value of 
and 
be positive real numbers satisfying 
and 
Prove that![\[\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.\]](http://latex.artofproblemsolving.com/0/a/b/0ab4b4e6c28a1ae67788525642661dbcd04d0bd8.png)
and its cyclic counterparts. Then by Holder's,![\[\sum_{cyc}{\frac{a}{(b+c-a)^2}}\sum_{cyc}a^2(b+c-a)\sum_{cyc}a^3(b+c-a)\geq (\sum_{cyc}a^2)^3=27\]](http://latex.artofproblemsolving.com/a/3/c/a3cb6da791767a4e300c16715ca205a358b155cc.png)
![\[\sum_{cyc}a^2(b+c-a)\leq 3abc
\]](http://latex.artofproblemsolving.com/c/1/0/c108aa9f3d3c1db3095b3398ae70afb99e40a936.png)
and![\[\sum_{cyc}a^3(b+c-a)\leq abc(a+b+c)
\]](http://latex.artofproblemsolving.com/c/f/c/cfc78867dc18cc01ad28f7c738d4143b44cd29f3.png)
Therefore![\[\sum_{cyc}{\frac{a}{(b+c-a)^2}}\geq \frac{9}{(abc)^2(a+b+c)}\geq\frac{3}{(abc)^2}\]](http://latex.artofproblemsolving.com/c/2/b/c2b597594083a310ca3e03a1623414369c7e9b5b.png)
and 
.
.
. then 2
>= 
. so, 
>. now , clearly , 
[if not , assume that 
. then 
. so, 
, which contradicts 
=3]. so , x,y,z are the sides of a triangle.
=3 and then applying LM method is sufficient . [here we must keep in mind the condition min(p+q,q+r,r+p) >
for choosing
are sides of triangle, so that 
with 
(which is shown by someone else).![$[x^{2}y^{2}(x+y)+y^{2}z^{2}(y+z)+z^{2}x^{2}(z+x)]\ge 12x^{2}y^{2}z^{2}$](http://latex.artofproblemsolving.com/8/f/f/8fff5ee1d64acd356131339246311f1a3792fcfa.png)
and homogenize the expression in (2) by multiplying LHS by 
and RHS by 
and ... then 2 >= . so, >. now , clearly , [if not , assume that . then . so, , which contradicts =3]. so , x,y,z are the sides of a triangle.=3 and then applying LM method is sufficient . [here we must keep in mind the condition min(p+q,q+r,r+p) > for choosing
and its cyclic counterparts. Then by Holder's,
and be positive real numbers satisfying and Prove that
and 
are positive. Thus by Hölder: 
Since by Schur: 
Thus the inequality of the problem holds.
. Now we can use 
lemma :
.And it is easy by Holder .
denote the left hand side. We know that by Schur's Inequality,
Since 
by Power Mean Inequality, we have![\[3abc \ge a^{2.5}(b+c-a) + b^{2.5}(c+a-b)+c^{2.5}(a+b-c)\]](http://latex.artofproblemsolving.com/0/5/4/054364705b54eae46771992a608a32f7255f0e9c.png)
By Holder's Inequality, 
so the desired holds.
we get 
thus, 
satisfy the triangle inequality. Let 
.![\[\sum{x}\sum{\frac{1}{x^2}}-\sum{\frac{1}{x}}=\sum{\frac{y+z}{x^2}}\overset{?}{\geq} \frac{384}{(y+z)^2(x+z)^2(x+y)^2}\]](http://latex.artofproblemsolving.com/b/d/e/bde6cae05f45b5f7185403a163dc29ba17639a32.png)
Let 
where 
.![\[3u.\frac{9v^4-6uw^3}{w^6}-\frac{3v^2}{w^3}\overset{?}{\geq} \frac{384}{(9uv^2-w^3)^2}\iff \frac{u(9v^4-6uw^3)}{w^6}\overset{?}{\geq} \frac{v^2}{w^3}+\frac{128}{(9uv^2-w^3)^2}\]](http://latex.artofproblemsolving.com/1/4/e/14e581127eba7f2aa75c9ac960841fb38539681c.png)
Note that 
holds and 
since 
.![\[\frac{u(9v^4-6uw^3)}{w^6}\geq \frac{3uv^4}{w^6}\geq \frac{3u^2}{w^3}\geq \frac{v^2}{w^3}+\frac{2u^2}{w^3}\geq \frac{v^2}{w^3}+\frac{2}{u^2v^4}\geq \frac{v^2}{w^3}+\frac{128}{(9uv^2-w^3)^2}\]](http://latex.artofproblemsolving.com/6/a/2/6a210f57cc322e628f1861efa44a0c96a6076faa.png)
As desired.
such that 
.
:
,
.
,
,
,![\[
\left(\frac{\sqrt{a}}{b+c-a},\frac{\sqrt{b}}{c+a-b},\frac{\sqrt{c}}{a+b-c}\right),
\qquad
\left(a^2\sqrt{b+c-a},b^2\sqrt{c+a-b},c^2\sqrt{a+b-c}\right)
\]](http://latex.artofproblemsolving.com/2/8/4/284c9e052ef49f5e1d6d1497a502da582512eb30.png)
![\[
a^4(b+c-a)+b^4(c+a-b)+c^4(a+b-c)\leq(a^2+b^2+c^2)abc=3abc
\]](http://latex.artofproblemsolving.com/7/a/6/7a68d3108737ce9be2c355b3d1893be41433d81c.png)
![\[
\frac{a^2\sqrt{a}}{\sqrt{b+c-a}}+\frac{b^2\sqrt{b}}{\sqrt{c+a-b}}+\frac{c^2\sqrt{c}}{\sqrt{a+b-c}}
\geq\frac{3}{\sqrt{abc}}.
\]](http://latex.artofproblemsolving.com/6/e/0/6e0f8d9d5cd792c0a2250c96dabc6d1ee70648c4.png)
is clearly convex for all positive real 
and weights 
,![\[
a^2\sqrt{a}(b+c-a)+b^2\sqrt{b}(c+a-b)+c^2\sqrt{c}(a+b-c)\leq
abc(\sqrt{a}+\sqrt{b}+\sqrt{c}),
\]](http://latex.artofproblemsolving.com/f/9/7/f97fdc462f0e39c3282100ad58aad366c4065970.png)
![\[
\left(\frac{a^2\sqrt{a}+b^2\sqrt{b}+c^2\sqrt{c}}{3}\right)^{\frac{3}{2}}
\geq\sqrt{\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}},
\]](http://latex.artofproblemsolving.com/b/4/a/b4a7f4db06f67e2ee17221eb3f4ab6f4521c017a.png)
.
, This is what I wrote that time :
which can be proved by Holder :
and etc 
?
and similarly for its cyclic counterparts. We write 
,
are positive. Our constraint becomes ![\[x^2+y^2+z^2 + xy+yz+zx=6.\]](http://latex.artofproblemsolving.com/a/f/5/af5b7948f2afbff6d0e44bdf8867b2c16d0932bc.png)
We need to show that ![\[\frac{y+z}{x^2} + \frac{z+x}{y^2} + \frac{x+y}{z^2} \ge \frac{6 \cdot 8^2}{(x+y)^2(y+z)^2(z+x)^2}.\]](http://latex.artofproblemsolving.com/f/f/6/ff61155e5ad23c1481401c732e4d44e63641c733.png)
Writing 
,
.![\[\left(\frac{AB-C}{C}\right)^2 (AB^2-2A^2C-BC) \ge 6 \cdot 8^2.\]](http://latex.artofproblemsolving.com/e/4/d/e4db0bca9c3e3dc52fabdf2d3c2fafe9696ae513.png)
Notice that the LHS is a decreasing function of 
.![\[\frac{2y}{x^2} + 2\frac{x+y}{y^2} \ge \frac{6 \cdot 8^2}{(x+y)^4 4y^2}\]](http://latex.artofproblemsolving.com/9/2/d/92dc419e3516122b185c09f5d1efca553af366b2.png)
with constraint ![\[x^2 + 2xy + 3y^2 = 6.\]](http://latex.artofproblemsolving.com/1/5/0/15019f852fc1728948824df7454a8f9ec60784e4.png)
The desired inequality rearranges to ![\[(x^3+x^2y+y^3)(x+y)^4 \ge 48x^2.\]](http://latex.artofproblemsolving.com/7/9/a/79ac7d6ea4daf67556bd38d6cfb6d1539f7dd92c.png)
to denote the polynomial 
.
and our desired inequality is ![\[ (\{ 1,1,0,1\})(\{1,4,6,4,1\}) \ge 48x^2. \]](http://latex.artofproblemsolving.com/2/a/5/2a507e7d5d1989de98c94fe9a16c261095564aa7.png)
Squaring both sides yields ![\[(\{1,2,1,2,2,0,1\})(\{1,8,28,56,70,56,28,8,1\}) \ge 48^2x^4.\]](http://latex.artofproblemsolving.com/c/d/7/cd7abb6ab9b687b5d52579e12789bc4ec18ee444.png)
Expanding the LHS gives ![\[\{1,10,45,122,228,324,379,380,325,234,143,74,30,8,1\} \ge 48^2x^4.\]](http://latex.artofproblemsolving.com/7/e/a/7ea1e1292834bff3b936732836109850612abfbd.png)
![\[\frac{48^2}{6^5}x^4(\{1,2,3\})^5\]](http://latex.artofproblemsolving.com/5/e/f/5ef0f2836899e693fbdefaad23afff5a0d650b7b.png)
using our constraint. This is equal to ![\[\frac{8}{27}x^4(\{1,2,3\})^5.\]](http://latex.artofproblemsolving.com/9/b/6/9b6149f4dc057e9c613bc04604dac0131a40b53d.png)
.
and 
.![\[(\{1,2,3\})^5 = (\{1,2,3\})(\{1,4,10,12,9\})^2 = \{1,10,55,200,530,1052,1590,1800,1485,810,243\}.\]](http://latex.artofproblemsolving.com/d/2/c/d2c4acaea5d0dcf4baaa6c4b095b560e89caec65.png)
![\[27(\{1,10,45,122,228,324,379,380,325,234,143,74,30,8,1\}) \ge 8(\{1,10,55,200,530,1052,1590,1800,1485,810,243,0,0,0,0\}).\]](http://latex.artofproblemsolving.com/b/f/7/bf7b68674a3fd838b4483f67f5cd3ace5d35bb0f.png)
This becomes ![\[\{27,270,1215,3294,6156,8748,10233,10260,8775,6318,3861,1998,810,216,27\} \ge \{8,80,440,1600,4240,8416,12720,14400,11880,6480,1944,0,0,0,0\}.\]](http://latex.artofproblemsolving.com/7/4/a/74af9b540a7d1ebdfb99bdde31953590da709fed.png)
Moving all terms to the LHS: ![\[\{19,190,775,1694,1916,332,-2487,-4140,-3105,-162,1917,1998,810,216,27\} \ge 0.\]](http://latex.artofproblemsolving.com/e/4/6/e46245f8313b30dea2aac3cbb1ea497c1cf88336.png)
This factors as ![\[(\{1,-1\})^2(\{19,228,1212,3890,8484,13410,15849,14148,9342,4374,1323,270,27\}) \ge 0,\]](http://latex.artofproblemsolving.com/1/6/e/16e03e345fae02899795bbe87e034c7d84ec52fd.png)
as desired. QED
?