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.
where 
is an odd integer. Let 
be a function defined on 
taking values in 
such that
for all 
for all 
; b) 
identical-looking coins in a circle. It is known that exactly two of them are fake. Real coins weigh the same, fake ones too, but they are lighter than the real ones. How can you determine in three weighings on a cup scale without weights whether there are fake coins lying nearby or not??
points of general position marked on the coordinate plane (i.e., points among which there are no three lying on the same straight line). Is there a polynomial of two variables 
a) of degree 
; b) of degree such that it equals to zero exactly at these marked points?
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
. Prove that 
. (
is area of triangle 
).
such that there exist integers 
with ![\[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]](http://latex.artofproblemsolving.com/e/f/a/efaa98e77ac0273284ee3446e4335d8cc4f681cb.png)
and 
doubloons, where and are given mutually prime natural numbers, with 
. A non-negative integer is called lucky if the sum in doubloons can be scored with using no more than 
coins. Find the number of lucky numbers.
with no parallel sides is inscribed in a circle 
. Circles 
are inscribed in triangles 
, respectively. Common external tangents are drawn between 
and 
, and 
, and 
, and and , not containing any sides of quadrilateral . A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle 
. Prove that the lines joining the centers of and , and , and the centers of and all intersect at one point.
be a quadrilateral inscribed in a circle 
such that 
. Let 
and 
be the midpoints of 
and 
respectively. The line 
meets again at 
. Prove that the tangent at to , the line and the line 
are concurrent.
be a continuous function. A chord is defined as a segment of integer length, parallel to the x-axis, whose endpoints lie on the graph of 
. It is known that the graph of contains exactly 
chords, one of which has length 2025. Find the minimum possible value of .
has been partitioned into disjoint pairs 
(
) so that for all 
, 
equals 
or 
. Prove that the sum ![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
be an acute-angled triangle with circumcircle 
. A circle is internally tangent to at 
and also tangent to 
at 
. Let and 
intersect at 
and 
respectively. Let and be points on line such that 
is the midpoint of 
and 
is the midpoint of 
. Lines 
and 
meet at 
and intersect again at 
and 
respectively. The ray 
meets the circumcircle of triangle 
again at 
.
. be a triangle and 
its circumcentre. A line tangent to the circumcircle of the triangle 
intersects sides at and at . Let 
be the image of under 
. Prove that the circumcircle of the triangle 
is tangent to the circumcircle of triangle .
be a semicircle with diameter and midpoint . Let be a point on different from and .
touches in a point , the segment 
in a point , and additionally the segment . The circle 
touches in a point and additionally the segments 
and .
and 
are perpendicular.
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.
and 
Prove 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
?