and 
Prove that
,
,
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
and 
are given. Let 
be the midpoint of the segment joining their centers, 
,
be arbitrary points on 
.
.
of points in the plane is balanced if, for any two different points 
and 
in 
in 
.
in 
.
,
points.
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 
)
.
