,
and 
are taken such that the point 
and 
.
and 
,
and 
are taken so that 
and 
.
.
and 
,
.
and 
be intersection points of lines 
and 
with circumcircle 
of triangle 
,
and 
intersection points of lines 
and 
with side 
.
,
with 
such that 
divides 
for all positive integers 
and 
with 
.
such that 
Follows for all reals 
.
denote the set of positive integers. Find all functions 
such that for positive integers 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
,
so that 
and 
.
so that 
.
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
Y by langkhach11112, rmtf1111, Adventure10, Mango247
Let 
be positive real numbers, 
. Denote by 
their geometric mean, and by 
the sequence of arithmetic means defined by ![\[ A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \]](//latex.artofproblemsolving.com/6/f/4/6f49684d75b0c4d0055613f1cf6bcafb034de25f.png)
Let 
be the geometric mean of . Prove the inequality ![\[
n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \]](//latex.artofproblemsolving.com/8/6/4/8644f80b9729df06edaf0de781fc98ae16392151.png)
and establish the cases of equality.
Proposed by Finbarr Holland, Ireland
be positive real numbers, 
. Denote by 
their geometric mean, and by 
the sequence of arithmetic means defined by ![\[ A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \]](http://latex.artofproblemsolving.com/6/f/4/6f49684d75b0c4d0055613f1cf6bcafb034de25f.png)
Let 
be the geometric mean of . Prove the inequality ![\[
n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \]](http://latex.artofproblemsolving.com/8/6/4/8644f80b9729df06edaf0de781fc98ae16392151.png)
and establish the cases of equality.Proposed by Finbarr Holland, Ireland
.
![\[m_k=\frac {A_{k-1}} {A_k}\]](http://latex.artofproblemsolving.com/2/5/0/250cccd8421871f38845ad1fa626910c53f7cd5d.png)
(suppose that 
)![\[\frac {a_k}{A_k}=\frac{kA_k-(k-1)A_{k-1}}{A_k}=k-(k-1)m_k\]](http://latex.artofproblemsolving.com/7/a/e/7ae77792a490a23b228d299b0cc02e972bdb71ba.png)
We have:![\[\sqrt[n]{\frac {G_n}{A_n}}=\sqrt[n^2]{m_2m_3^2\dots m_n^{n-1}}\]](http://latex.artofproblemsolving.com/9/7/c/97c820ef4f6beafd339730a0e75c35eb2b8f8387.png)
![\[\frac {g_n}{G_n}=\sqrt[n]{\prod_{k=1}^n(k-(k-1)m_k)}\]](http://latex.artofproblemsolving.com/9/3/1/931093b301839f44799c08e643cb10f0126bc49f.png)
![\[n\sqrt[n^2]{1^{\frac{n(n+1)}2}m_2m_3^2\dots m_n^{n-1}}\leqslant \frac{\frac{n(n+1)} 2+\sum_{k=2}^n(k-1)m_k}n=\frac{n+1}2+\frac {\sum_{k=1}^n(k-1)m_k}n\]](http://latex.artofproblemsolving.com/f/9/f/f9fefcaa882cd774de347e629b1a508444cba4ca.png)
![\[\sqrt[n]{\prod_{k=1}^n(k-(k-1)m_k)}\leqslant \frac{n+1} 2-\frac {\sum_{k=1}^n(k-1)m_k}n\]](http://latex.artofproblemsolving.com/e/4/7/e47ac65843e8f978ad273064b31b3663dc05a6e5.png)
.
![\[ n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}+\sqrt[n]{(2-A_1/A_2)...(n-(n-1)A_{n-1}/A_n)}\leq n+1. \]](http://latex.artofproblemsolving.com/3/3/6/336415f6eafe083c51d1b2cff000cdfecd322541.png)
Use the AM-GM inequality to deduce that the second term of the LHS is at most ![\[ \frac{1+(2-A_1/A_2)+...+(n-(n-1)A_{n-1}/A_n)}{n}. \]](http://latex.artofproblemsolving.com/7/8/c/78cbc35bf92e8204655222f62febe2664befd303.png)
![\[ \frac{n+1}{2}+\frac{A_1/A_2+...+(n-1)A_{n-1}/A_n}{n} \geq n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}. \]](http://latex.artofproblemsolving.com/4/3/0/430778ca3508ef6a1e3c63d3a32e68570e670006.png)
Now, apply generalized AM-GM inequality to deduce that the LHS is at least ![$ (n+1)/2+(n-1)/2\cdot\sqrt[(n^2-n)/2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}$](http://latex.artofproblemsolving.com/9/c/c/9cc2690fb897b30f4b1858f1b8e4df3afac50b3d.png)
.
.![\[n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}+\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le n+1.\]](http://latex.artofproblemsolving.com/f/5/2/f522c6ea028d1d84971e84022e8489428097656c.png)
We see that ![\[n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}\le \frac{n-1}{n}c_{n-1}+\cdots+\frac{1}{n}c_1+\frac{n+1}{2},\]](http://latex.artofproblemsolving.com/4/2/d/42ddc5b8de2f45d64792b221940d888fcf10e386.png)
and ![\[\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le \frac{n+1}{2}-\frac{1}{n}c_1-\cdots-\frac{n-1}{n}c_{n-1},\]](http://latex.artofproblemsolving.com/f/b/7/fb715fb3c029681e046434d1709c3e70986a981a.png)
so adding gives the desired estimate.
case, which was solved with a similar AM-GM as above.
.
for all ![\begin{align*}
n \sqrt[n]{\frac{G_n}{A_n}}+ \frac{g_n}{G_n} &= n \sqrt[n^2]{\frac{A_1A_2A_3\dots A_n}{A_n^n}} + \sqrt[n]{\frac{a_1}{A_1}\cdot \frac{a_2}{A_2}\cdot \ldots \cdot \frac{a_n}{A_n}} \\
&= n\sqrt[n^2]{\left(\frac{A_1}{A_n}\right)\left(\frac{A_2}{A_n}\right)\dots \left(\frac{A_{n-1}}{A_n}\right)} + \sqrt[n]{\left(\frac{2A_2-A_1}{A_2}\right)\left(\frac{3A_2-2A_2}{A_3}\right)\dots\left(\frac{nA_n-(n-1)A_{n-1}}{A_n}\right)} \\
&=n\sqrt[n^2]{1^{\frac{n(n+1)}{2}}b_1 b_2^2b_3^3\dots b_{n-1}^{n-1}}+\sqrt[n]{(1)(2-b_1)(3-2b_2)(4-3b_3)\dots(n-(n-1)b_{n-1})} \\
&\le n\cdot \frac{\frac{n(n+1)}{2}+b_1+2b_2+\dots + (n-1)b_{n-1}}{n^2} + \frac{\frac{n(n+1)}{2}-b_1-2b_2-\dots-(b-1)b_{n-1}}{n} \\
= n+1
\end{align*}](http://latex.artofproblemsolving.com/6/7/8/6789798b46a6991fcb3ce5c8dd17927afef94ff7.png)
as desired.![$n \cdot \sqrt[n^2]{\frac{A_1 \cdots A_{n-1}}{A_n^{n-1}}}+\sqrt[n]{\left(2-\frac{A_1}{A_2} \right) \cdots \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)} \le n+1$](http://latex.artofproblemsolving.com/f/7/a/f7af46bd025da8f144af9c513b4fb9b5eac8bbce.png)
.![\[ \frac{n+1}{2}-\frac{\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n}=\frac{1+\left(2-\frac{A_1}{A_2} \right)+ \cdots + \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)}{n} \ge \sqrt[n]{\left(2-\frac{A_1}{A_2} \right) \cdots \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)} \]](http://latex.artofproblemsolving.com/7/3/c/73c35d0a20fd18651a965f185b1efd04bce1deba.png)
Now now all we need to prove after replacing this, is the following:![\[ \frac{n+1}{2}+\frac{\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n} \ge n \cdot \sqrt[n^2]{\frac{A_1 \cdots A_{n-1}}{A_n^{n-1}}} \]](http://latex.artofproblemsolving.com/a/a/d/aada3aa153ddf2ee26a0318eeec1a98aede10777.png)
But from AM-GM once again one has that:![\[ \frac{\frac{n(n+1)}{2}+\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n^2} \ge \sqrt[n^2]{\frac{1^{\frac{n(n+1)}{2}} \cdot A_1 \cdots A_{n-1}}{A_n^{n-1}}} \]](http://latex.artofproblemsolving.com/8/4/2/842a726eb06971966959c55a4913297ff50f14d0.png)
(And thus eq case happens when 
)
.
