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 the set of black cells at the end of the game. Ali and Shayan respectively aim to minimize and maximise the perimeter of the shape by playing optimally. (The perimeter of shape is defined as the total length of the boundary segments between a black and a white cell.), assuming both players play optimally?
+1 w
be a positive integer and let 
be an irrational number in 
. Show that for every integer 
, there exists a positive integer 
such that 
.
friends, people are seated around a circular table in a fixed order. Ali places apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction). such that after some number of steps, the situation reaches a point where each person has exactly one apple.
is a family of non-empty subsets of , such that any two distinct subsets in the family are disjoint, and the union of all subsets equals . We say that a partition of a set of integers 
is separated if each subset in the partition does not contain consecutive integers. Prove that, for every positive integer 
, the number of partitions of the set 
is equal to the number of separated partitions of the set 
.
is a separated partition of the set 
. On the other hand, 
is a partition of the same set, but it is not separated since 
contains consecutive integers.
be odd integers such that 
and 
. Prove that if 
and 
for some integers 
and 
, then 
.
such that 
Prove that
Where 
such that 
Prove that
parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on is not greater than 
?
be reals such that 
and 
Prove that
Let be reals such that 
and Prove that![$$ a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$$](http://latex.artofproblemsolving.com/6/2/2/62211222c148c86d65526d86dde603cb6598925f.png)
![$$ a^2+2b^2+c^2\geq 2\sqrt[3]{4} $$](http://latex.artofproblemsolving.com/3/8/3/383d38cc87c1b1ea1e17e10e4bb95bed79329786.png)
be the set of positive integers. Find all functions 
, defined on 
and taking values in , such that 
for every positive integer .
is used in some steps. Some steps also use Muirhead's
(We're motivated to do this because the fifth power is really out of place.) Indeed, note that 
which is obviously true. 
such that 
Where 
then find minimal value of 
Prove that
Where 
and 
.
such that: 
.
be three positive reals such that 
.![\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]](http://latex.artofproblemsolving.com/a/1/4/a14adf0f1e35df57c734c7df701d5a67da22dc3c.png)
.
,
,
,
,
.![\[ 3 - \left( \sum { \ell^2 x'^2 } \right) \cdot \left ( \sum \frac 1 { \ell^5 x'^5 + \ell^2 ( y'^2+z'^2 ) } \right) \geq 0 . \]](http://latex.artofproblemsolving.com/5/2/9/5296924f9989e72fa51991016641f4fee4af2227.png)
,
.
is a positive constant. It is easy to see that each of the functions 
is decreasing, so 
.
,
,
.![\[ 3 \geq \left( \sum \frac { b^2c^2 }{a^4 } \right) \cdot \left( \sum \frac 1 { \displaystyle \frac {b^5c^5}{a^{10}} + \frac {c^2a^2}{b^4} + \frac {a^2b^2}{c^4} } \right) \Leftrightarrow \]](http://latex.artofproblemsolving.com/6/d/2/6d2ed1031bcd229b907ac963e411f5ffb579276e.png)
![\[ 3 \geq \frac 1{a^4b^4c^4} \left( \sum {b^6c^6} \right) \cdot \left( \sum \frac { a^{10}b^4c^4 }{ b^9c^9 + c^6a^{12} + b^6 a^{12} } \right) \ \Leftrightarrow \]](http://latex.artofproblemsolving.com/f/a/0/fa02b872d3cb90e9333dbec66581762505f954e0.png)
![\[ 3 \geq \left( \sum b^6 c^6 \right) \cdot \left( \sum \frac { a^6 }{ b^9c^9 + c^6a^{12} + b^6 a^{12} } \right) \ \Leftrightarrow \]](http://latex.artofproblemsolving.com/7/d/0/7d05d880cf2f058ac667b5c864411db535b15c6f.png)
![\[ 3 \geq ( n^2p^2 + p^2 m^2 + m^2n^2 ) \cdot \left( \sum \frac {m^2}{n^3p^3 + p^2m^4 + n^2 m^4 } \right) , \]](http://latex.artofproblemsolving.com/f/5/0/f509296683a3741f1c20ab301b98f6e13d7f7dbf.png)
with 
respectively.![\[ (n^3p^3 + p^2m^4 + n^2 m^4 )( np + p^2 + n^2 ) \geq ( n^2p^2 + p^2 m^2 + m^2n^2 )^2 \]](http://latex.artofproblemsolving.com/e/9/a/e9a2ae2f5808bda7eba9d781ce0f1ae1314a0fc9.png)
from Cauchy. Using this minoration we have to prove that![\[ 3 \geq \frac { \sum ( m^2np + m^2p^2 + m^2n^2 ) } { \sum n^2p^2 } \]](http://latex.artofproblemsolving.com/7/c/2/7c2101c7f5c30cd2dd4aca6e007d2d95cb4ec7a0.png)
which is obviously true as ![\[\sum n^2p^2 \geq \sum (mn)(mp) \Leftrightarrow \frac 12 \left ( \sum (mn-mp)^2 \right) \geq 0 . \]](http://latex.artofproblemsolving.com/2/6/2/2620f2af6d13e101ade43b845370a03511512f9e.png)
.
holds. At first, we may assume that 
by the same positive constant 
,
increases. Then we have 
.
,
etc, 
.
is from Yugoslavian IMO TST this year, as I remember.
![\[ \sum_{cyc} \frac{x^5}{x^5+y^2+z^2} \geq \sum_{cyc} \frac{x^2}{x^5+y^2+z^2} \]](http://latex.artofproblemsolving.com/b/0/3/b03e90423251319c86f3bbda5e1587ba5680f222.png)
![\[ \sum_{cyc} \left( 1 - \frac{y^2+z^2}{x^5+y^2+z^2} \right) \geq \sum_{cyc} \frac{x^2}{x^5+y^2+z^2} \]](http://latex.artofproblemsolving.com/5/7/8/578f6a19b181e38c6e4bd249d08488cd0ee1a693.png)
![\[ 3 \geq (x^2+y^2+z^2)\cdot \left( \sum_{cyc} \frac{1}{x^5+y^2+z^2} \right) \]](http://latex.artofproblemsolving.com/1/1/0/11046817c5f7f929ef10ca1485e4b655229b812b.png)
![\[ 3 \geq (a^3bc+ab^3c+abc^3)( \frac{1}{a^5+ab^3c+abc^3} + \frac{1}{a^3bc+b^5+abc^3} + \frac{1}{a^3bc+ab^3c+c^5} )\]](http://latex.artofproblemsolving.com/5/f/9/5f96ac006e2652f4987a7b78cf1dab1de96f0951.png)
![\[ \frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{x^{2}+y^{5}+z^{2}}+\frac{z^{5}-z^{2}}{x^{2}+y^{2}+z^{5}}\geq 0\]](http://latex.artofproblemsolving.com/f/b/6/fb66a2213567bfc1bc78b828b18fb584b09126ed.png)
is equivalent to : ![\[ \frac{1}{x^{5}+y^{2}+z^{2}}+\frac{1}{x^{2}+y^{5}+z^{2}}+\frac{1}{x^{2}+y^{2}+z^{5}}\leq \frac{3}{x^{2}+y^{2}+z^{2}}\]](http://latex.artofproblemsolving.com/0/f/b/0fbeabfd759aca968327387b6e0031f9111fa98f.png)
Now, from Cauchy's inequality we have: ![\[( x^{2}\cdot x^{3}+y^{2}+z^{2}) ( x^{2}\cdot \frac{1}{x^{3}}+y^{2}+z^{2}) \geq (x^{2}+y^{2}+z^{2})^{2}\]](http://latex.artofproblemsolving.com/8/8/c/88c41bd51a78615aa6f886ee28ca8fd97c167179.png)
But from the given condition we have 
so we obtain that: ![\[(x^{5}+y^{2}+z^{2}) (yz+y^{2}+z^{2}) \geq ( x^{2}\cdot x^{3}+y^{2}+z^{2}) ( x^{2}\cdot \frac{1}{x^{3}}+y^{2}+z^{2}) \geq (x^{2}+y^{2}+z^{2})^{2}\]](http://latex.artofproblemsolving.com/9/2/6/926b36e6ae8f0c5c5bea5d5705bf724f07f6b818.png)
All that is left now is to notice that ![\[ \sum_{\textrm{cyc}}\frac{1}{x^{5}+y^{2}+z^{2}}\leq \sum_{\textrm{cyc}}\frac{yz+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}\leq \sum_{\textrm{cyc}}\frac{\frac{y^{2}+z^{2}}{2}+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}= \frac{3}{x^{2}+y^{2}+z^{2}}\]](http://latex.artofproblemsolving.com/2/9/7/2973c9ee7f676b9713314f4a3f88962421815113.png)
, you beat me to the bunch.
bottom is the key, and a nice trick. I was able to see this trick fairly easily, but I have to admit that I had some recent inspiration.
and use some obvious reductions:
is what we want, or equivalently
,
where 
and likewise 
.
if we can prove that 
,
.
,
.
where 
.
,
]
goes to infinity may not be concave.
we get:![\[ \sum_{cyc}\frac{1}{x^{5}+y^{2}+z^{2}}\leq \sum_{cyc}\frac{yz+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}\leq \sum_{cyc}\frac{\frac{y^{2}+z^{2}}{2}+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}= \frac{3}{x^{2}+y^{2}+z^{2}}\]](http://latex.artofproblemsolving.com/f/4/2/f42e36c0efb88a065b87b695ed014e29464155b1.png)
![\[(x^5+y^2+z^2)(yz+y^2+z^2) \ge (x^2 \sqrt{xyz}+y^2+z^2)^2 \ge (x^2+y^2+z^2)^2,\]](http://latex.artofproblemsolving.com/b/9/b/b9b655d15d24b6aab6c7dfdc178b5fc58adfe53e.png)
![\[\sum_{\text{cyc}} \frac{1}{x^5+y^2+z^2} \leq \sum_{\text{cyc}} \frac{yz+y^2+z^2}{(x^2+y^2+z^2)^2} \leq \frac{3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2} = \sum_{\text{cyc}} \frac{1}{x^2+y^2+z^2}.\]](http://latex.artofproblemsolving.com/9/6/4/96450476a5759f7139492f107c5099583229433e.png)
,
which becomes 
after homogenizing. Now cross multiplying everything and simplifying gives ![\[\frac12\sum_{\text{sym}}x^8y^2z^2+2\sum_{\text{sym}}x^7y^5-\sum_{\text{sym}}x^6y^5z-\sum_{\text{sym}}x^6y^4z^2-\frac12\sum_{\text{sym}}x^5y^5z^2+\sum_{\text{cyc}}x^6yz(x^2-yz)^2\ge0,\]](http://latex.artofproblemsolving.com/1/d/e/1de17fc5cf66411ecb74e966c99b75067657190b.png)
which follows from Muirhead.
So, 
.
and assume wlog that 
.
then we want to prove 
.
which is obviously true.
From cauchy
so the following suffices:
But this is trivial by 
