?
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
be six distinct real numbers satisfying the following condition:
,![\[
x_i^6 + (-1)^{i+1} x_i = -x_1 x_2 x_3 x_4 x_5 x_6.
\]](http://latex.artofproblemsolving.com/3/6/f/36f886654cf5f7af956e2475be94acbe62bb8153.png)
.
,
integers and 
is a positive integer. Suppose that the integers
satisfy: 
are divisible by k. Prove that:
is divisible by k
such that 
and
.
and 
Prove that![$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$](http://latex.artofproblemsolving.com/5/6/1/561a990e5885863c77d454ffa3f1973e1bbf60b4.png)
![$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$](http://latex.artofproblemsolving.com/4/b/1/4b14911672688250c909571d788aaf10724462eb.png)
Where 
numebr on his list? Write its last two digits as your answer.
grid of square tiles. Robert chooses four of these squares uniformly at random. The probability that the centers of these four squares form a square themselves is 
where 
are coprime naturals. Write the largest 
digit odd factor of 
.
.
which touches one face of the cube and passes through all 
vertices of the face opposite to it. Find the integer closest to 
are distinct numbers in AP such that 
are also in AP. Find the smallest possible value of 
(in degrees).
and a plane 
,
and 
.
.
and 
,
denote the increasing arithmetic sequence of integers whose first term is 
and whose common difference is 
is the sequence 
.
?
is the set of positive real solutions to the system![\[
x^3 + y^3 + z^3 = x + y + z
\]](http://latex.artofproblemsolving.com/7/7/a/77a748a8b55a6b44761bced9bb1031193b6127dc.png)
![\[
x^2 + y^2 + z^2 = xyz
\]](http://latex.artofproblemsolving.com/9/3/a/93a4dffc6fd17b9f19f738d3b3865f56aa0151a9.png)
then find the number of elements in 
Y by Amir Hossein, Hydrogen-Helium, Davi-8191, tenplusten, Vietjung, OlympusHero, itslumi, Karangy, centslordm, mathematicsy, jhu08, megarnie, ehuseyinyigit, Adventure10, Mango247, GeoKing, ohiorizzler1434, and 1 other user
Let 
be an integer, and let 
be positive real numbers such that 
. Prove that
![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](//latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
Proposed by Angelo Di Pasquale, Australia
be an integer, and let 
be positive real numbers such that 
. Prove that![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](http://latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
Proposed by Angelo Di Pasquale, Australia
This post has been edited 2 times. Last edited by delegat, Jul 11, 2012, 11:41 AM
![$(a_k+1)=\left(a_k+\frac 1{k-1}+\cdots+\frac 1{k-1}\right)\geq k\sqrt[k]{\frac{a_k}{(k-1)^{k-1}}};$](http://latex.artofproblemsolving.com/8/f/d/8fdb1ed6783f2ee120e7ff4861fe2020266c5131.png)
to 
we see that![\[\prod_{k=2}^n(a_k+1)^k\geq n^na_2a_3\cdots a_{n}=n^n.\]](http://latex.artofproblemsolving.com/c/d/b/cdb3cb3fa852f6d6ba306a2c93782a08ab84c5f7.png)
for all 
which is not possible. 
,
such that 
for all positive 
.
and so we have 
for 
.
to get the optimal 
and we are done.
also appears, for example, in lorincz's solution to this Russia 2007 problem:
,
.![\[ (x_2 +x_3)^2 (x_3 +x_4)^3 ... (x_n +x_2)^n > n^n x_3 ^2 x_4 ^3 ... x_{n}^{n-1} x_2 ^n \]](http://latex.artofproblemsolving.com/4/5/c/45c140905ec49dbc491644f9affbf68c9a31fd98.png)
.
,
given by 
,
,
reaches its minimum at 
.
yields (by telescoping) that LHS is at least 
,
for each 
.
for all k, then 
![\begin{align*}
1+a_2 &\ge 2\sqrt{a_2} \\
\frac{1}{2}+\frac{1}{2}+a_3 &\ge 3\sqrt[3]{\left(\frac{1}{2}\right)^2a_3} \\
\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+a_4 &\ge 4\sqrt[4]{\left(\frac{1}{3}\right)^3a_4} \\
&\; \; \vdots \\
\frac{1}{n-1}+\cdots + \frac{1}{n-1} + a_n &\ge n\sqrt[n]{\left(\frac{1}{n-1}\right)^{n-1}a_n}
\end{align*}](http://latex.artofproblemsolving.com/c/6/7/c67a3d26a760f3f47af7508dd90e59874e3db7b9.png)
Taking the 
t
t![\[(1+a_2)^2 (1+a_3)^3 \dots (1+a_n)^n \ge n^n\]](http://latex.artofproblemsolving.com/d/6/9/d69230220ad98bd0265399990e20ea53a1a93501.png)
Equality occurs when the individual equalities for each inequality occurs. Equality for AM-GM occurs when all the terms are equal. This is when![\[a_i = \frac{1}{i-1}\]](http://latex.artofproblemsolving.com/6/9/f/69f12d25689cc2253f92d58f6a141e24c9e63fd9.png)
However, in this case, the product isn't 
Now, 
For equality to hold, we require 
but then 
which is never possible. Therefore, equality never holds, and 
Now here equality occurs when 
for all 
for 
which is not true thus equality cannot hold
or more generally ![$$(1+a_k)^k
= \left( \underbrace{\frac{1}{k-1} + \frac{1}{k-1} + \dots + \frac{1}{k-1}}_{k-1} + a_k \right)^k \geq \left( k \sqrt[k]{\left( \frac{1}{k-1} \right)^{k-1} a_k} \right) ^k = \frac{k^k}{(k-1)^{k-1}}a_k$$](http://latex.artofproblemsolving.com/0/d/7/0d7aec006fb13fa28a9e210205d3e1d4ca5cac94.png)
Now telescoping gives the desired result, because all the 
cancel out from the condition. Also equality cannot hold because for equality in the AM-GM's we need 
which does not satisfy our condition. Done
due to 
.
.
but in that case the product isn't 
![$$\ge k\sqrt[k]{\frac{a_k}{(k-1)^{k-1}}}$$](http://latex.artofproblemsolving.com/a/f/4/af4d77b62d638e7b2ee5cf6e3c697bf756024559.png)
by AM-GM. Therefore, 
Multiplying these expressions together, we know that 
Equality is impossible as that would imply 
for every 
(
Thus multiplying the equations yields 
as the product telescopes. Note that equality holds when 
Therefore, our inequality is strict so we have shown the desired result. QED
In such a term, after expanding and counting multiple like terms separately we will have 
terms. Then applying AM-GM we see that,![\begin{align*}
(1+a_n)^n &= a_n^n+na_n^{n-1}+\dots + 1\\
&\geq 2^n(\sqrt[2^n]{a_n^{k}})\\
&\geq 2^n(a_n^{\frac{k}{2^n}})\\
& > 2^na_n
\end{align*}](http://latex.artofproblemsolving.com/a/c/c/acc0e6c1758c65637507fe63b7cf00dcf52f85f3.png)
Here, notice that we obtain the fourth line from 
)
of the required inequality, 
This reduces the desired inequality to 
for all 
This inequality is obviously true for 
for all 
,
which was the required inequality.
substitution + induction that went nowhere :C
![\[\prod_{k=2}^{n}(1+a_k)^k=\prod_{k=2}^{n}\left((k-1)\cdot\frac{1}{k-1}+a_k\right)^k
\ge\prod_{k=2}^{n}\frac{k^k a_k}{(k-1)^{k-1}}=k^k.\]](http://latex.artofproblemsolving.com/5/1/b/51b78a211f24567fc3e6bee3827b1b403c75cd1c.png)
BUT:: EQUALITY CASE DOESNT WORK!!!!!!!!!!!! BECAUSE 
really helpful. Because this triggers us to use weighted AM-GM: ![$$\dfrac{(x-1) \cdot \dfrac{1}{x-1} + a_x}{x} \ge \sqrt[x]{\dfrac{a_x}{(x-1)^{x-1}}}.$$](http://latex.artofproblemsolving.com/d/a/e/daeaf2fc0ddebf11e0df818f892c59cfe6a3d951.png)
We look at this through induction (multiplying it out) to find the desired result, but with 
instead of 
.
for all 
,
By telescope, we hhave that 
However, equality occurs when 
Since 
is never an integer, we are done
