AoPS Academy
be an infinite sequence of positive integers. Prove that there exists a unique integer 
such that![\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]](http://latex.artofproblemsolving.com/9/4/f/94fc5a51b7e68588123c5b527fe75183bc4c4937.png)
+1 w
with 
, 
and 
for all positive integers 
.
,
if 
is a root of 
and solve the equation.
are constructed on diameter 
of a semicircle of radius 
. The centers of the small semicircles divide 
into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?
be an infinite geometric sequence where all the terms are positive integers. Amanda lists down the two-digit numbers formed by the last two digits of each term in the sequence in order, adding a leading zero if necessary. When she is about to write a two-digit number that she has already written before, she stops. For example, if the sequence was 
, Amanda will list the numbers 
, then stop.
and 
. If 
, how many possible values are there for 
?
regions of a three-set Venn Diagram with 
colors such that each color is used at least once? Two colorings are considered the same if one can be reached from the other by rotation and reflection.
be 
, and 
. What is the sum 
?
through 
, shuffle them, and then place them in a stack facedown, turning over the top card next to the stack. They then take turns either drawing the card at the top of the stack into their hand, showing the drawn card to the other player, or drawing the card that is faceup, replacing it with the card on the top of the pile. This is repeated until all cards are drawn, and the player with the largest sum for their cards wins. What is the probability that the player who goes second wins, assuming optimal play? such that 
is also prime. vertices of an octagon each red, black, and white, such that no two adjacent sides are the same color?
are all real positive number that: 
. Prove that:
, such that for 
![\[ \sum_{\text{cyc}} \left( \frac{x+y}{2(x+y+3z)} \right)^n \ge \frac{3}{5^n}. \]](http://latex.artofproblemsolving.com/6/4/b/64b56aa0977827ffae0e2e199d8ad3738cda5c6f.png)
Impose the condition 
, and let 
then 
, so by Jensen we must have that![\[ f(x) + f(y) + f(z) \ge 3f \left(\frac{x+y+z}{3} \right) = 3f \left( \frac 13 \right) = \frac{3}{5^n}, \]](http://latex.artofproblemsolving.com/0/5/b/05b1e39c090404f2f9961eac732cfdde79319f3f.png)
as desired.
Using Cauchy Schwartz inequality,
.
![$M\leq\min_{x>0}{\frac{x^3+1}{x^2}}=\frac3{\sqrt[3]4}.$](http://latex.artofproblemsolving.com/0/f/6/0f650ac11b1f8dbc980b1cdce6900c49e928b2df.png)
real positive numbers and 
and 
,
,
.
Since 
.
and 
.
![$x = \sqrt[n]{x_1x_2...x_n}$](http://latex.artofproblemsolving.com/9/5/0/950e616f846b5a4359023014bdac6d4bc813286e.png)
.
:
then 
.
:
.
Again by AM-GM inequality: 
,
,
Equality holds when 
.
.
Take the cyclic sum we have 
Need to prove 
or 
Equality occurs iff 
are three roots of 
.
.
.
prove that 
,
,
Moreover, by Cauchy's inequality,
Summing these two inequalities and dividing by 
gives us![\[\sum_{k=1}^nf(x_k)(x_k-x_{k-1})=A\ge\int_0^1f(x)\operatorname dx+\frac1{2n}\left(\int_0^1\sqrt{f^\prime(x)}\operatorname dx\right)^2.\]](http://latex.artofproblemsolving.com/3/4/5/3451e81916dfa491a76417ea355ed078a1fe5254.png)
and 
be two triangles and 
be an arbitrary point. Let 
be the inradius of triangle 
be the circumradius of triangle 
The problem was posted ![\[\textcolor[RGB]{11,83,148}{(PA+PB)(PB+PC)(PC+PA)\ge64r^3}.\]](http://latex.artofproblemsolving.com/1/e/9/1e93348377bf965dbf3f4217e826b75729f3f202.png)
Note that this is a generation of Euler's inequality ![$\textcolor[RGB]{11,83,148}{R\ge2r}$](http://latex.artofproblemsolving.com/3/6/a/36acaef398fa5cddca0d9efc1bc3ce58bdeac4ab.png)
.
where 
and 
For arbitary 
Determine the maximum value of 
and 
in the plane. For every line through the origin, let the projection of the vectors onto the line be 
and 
.![\[\sum_{i=1}^{n}|A_i| \ge \sum_{i=1}^{m}|B_i|.
\]](http://latex.artofproblemsolving.com/6/d/0/6d0ab9ed16be689155c0ce314012125286c0391f.png)
Show that we have ![\[\sum_{i=1}^{n}|a_i| \ge \sum_{i=1}^{m}|b_i|\]](http://latex.artofproblemsolving.com/5/0/c/50c9fef06e4588cc0a1fb01235c4fb63f50cfa01.png)
where 
is the length of the vector 
.
,
satisfy 
.
must hold (Here the indices are in modulo 
,
.
where 
is the projection length
Yongxi Wang
,
by 
and divide 
by the same amount. Then, the conditions still hold and the product 
increases. Hence, we can see that all the numbers at least 
.
of them that are 
numbers. These multiply to 
.
where 
.
by Jensen's inequality. It remains to prove that 
In other words, we wish to prove 
However, 
,
,
,
are complex numbers of length at most 
.
denote the area of the triangle formed by 
,
and 
.
.
.
be the point corresponding to 
denote the angle 
(![$\alpha_k\in [0,\pi]$](http://latex.artofproblemsolving.com/d/4/2/d42ea2459e5918fa1df28b23d65d24b4d5da39fd.png)
)
.
.
for some 
,
are distinct.
satisfies 
,
.
is at most 
.
to 
,
.
.
.
,
replaced by 
.
,
and 
share a common point. This completes the inductive step.
is concave on 
so 
meaning that 
.
be a set of positive integers with the following property: for every positive integer in 
.
.
belongs to 
cannot be 
.
is the sum of 
for all elements 
,
.
be the number of elements of 
digits. For any positive integer 
sequences of the first 
.
.
,
be integers such that 
.
.
,![\[\sum_{k=1}^{10}ka_k=\sum_{k=1}^{10}k(b_k+k-5)=110+\sum_{k=1}^{10}kb_k=111,\]](http://latex.artofproblemsolving.com/4/e/2/4e2e0d902138fe2f3eb7cb64f0af937be2ffe518.png)
thereby, 
.
,![\begin{align*}
S&:=\sum_{k=1}^{10}a_k^2+\left(\sum_{k=1}^{10}a_k\right)^2=\sum_{k=1}^{10}(b_k+k-5)^2+\left[\sum_{k=1}^{10}(b_k+k-5)\right]^2\\
&=\sum_{k=1}^{10}b_k^2+\sum_{k=1}^{10}(2k-10)b_k+85+\left(t+5\right)^2=\sum_{k=1}^{10}b_k^2-10t+87+(t+5)^2\\
&=\sum_{k=1}^{10}b_k^2+t^2+112\quad\overset{\text{C-S}}\ge\quad\frac{t^2}{10}+t^2+112.\tag1
\end{align*}](http://latex.artofproblemsolving.com/f/0/a/f0a69b55d79728865022035725bfda14ba9b6968.png)
If 
,
,
implies that 
.
,
for all 
and 
,
and 
.![\[S\ge b_i^2+b_j^2+t^2+112\ge114.\]](http://latex.artofproblemsolving.com/a/e/0/ae0b9756e0d14e059e562545b1d61a09a15bf4b0.png)
We conclude that 
because it can be obtained for 
,
.
,
,![$c\in[0,1]$](http://latex.artofproblemsolving.com/3/5/7/357d86e43515b8c7d9c6921417ae3d4e5d1c6518.png)
be such that 
.![\[1\le\frac a{1+2bc}+\frac b{1+2ca}+\frac c{1+2ab}\le2.\]](http://latex.artofproblemsolving.com/5/7/d/57d1b8ae8bc78dea457be1d09a84ce1ca2cb02a6.png)
,
,
.
.![\[\frac{2pq+2r+p}{8r^2+4rp+2q+1}.\]](http://latex.artofproblemsolving.com/1/d/3/1d3eb192469395ed5cb3a1190adce848efa23858.png)
Hence we just have to show 
as 
,
.
,
Therefore it is enough to show 
which is obvious from 
alone.
,
be nonnegative integers with sum 
,
be real numbers such that 
and 
for each 
,
.
,
and 
is at most 
.
,
into 
which increases the sum. Therefore, we can assume that 
are all nonnegative. 
is the area that four points; 
,
a
gon 
forms. However, since considering the slope, we can see that every 
,
,
with area ![$$ \sum_{i=1}^n(a_i+a_{i-1})b_i = 2[P_0P_1\cdots P_n] \le 2 $$](http://latex.artofproblemsolving.com/1/a/8/1a89f877791530808c8fa673e379f6175bdbd0c6.png)
such that for any 
,![\[ (x_1 + 2x_2 + \cdots + nx_n)(x_1^2 + x_2^2 + \cdots + x_n^2) \ge c (x_1 + x_2 + \cdots + x_n)^3 \]](http://latex.artofproblemsolving.com/1/e/0/1e03d09f0fab71446f6e22ac13bdf712320a53cf.png)
for all indices 
,![\[c\le\frac{\sum ia_i\sum a_i^2}{\left(\sum a_i\right)^3}=\frac{\frac{(n-1)n(n+1)}6\cdot\frac{(n-1)n(2n-1)}6}{\left(\frac{n(n-1)}2\right)^3}\to\frac49~(n\to+\infty).\]](http://latex.artofproblemsolving.com/b/6/7/b678c3d1e842e0e99d79d5c64a0ec923702c73c6.png)
We will then prove the inequality for 
.
,
Let 
.
where the last inequality follows from substituting 
and expansion. Hence,![\[\sum_{k=1}^na_k\le\frac1b\sum_{k=1}^nka_k+\sqrt{\textstyle\sum\limits_{k=1}^na_k^2}\cdot\sqrt{\frac b3}=\frac1b\sum_{k=1}^nka_k+2\sqrt{\textstyle\sum\limits_{k=1}^na_k^2}\cdot\sqrt{\frac b{12}}.\]](http://latex.artofproblemsolving.com/6/6/6/66685af70826d493627033619e4d5ae981d9acb7.png)
To obtain the best bound, we should let![$$\frac1b\sum_{k=1}^nka_k=\sqrt{\textstyle\sum\limits_{k=1}^na_k^2}\cdot\sqrt{\frac b{12}}\implies b=\sqrt[3]{\frac{12\left(\sum ka_k\right)^2}{\sum a_k^2}}.$$](http://latex.artofproblemsolving.com/f/5/6/f56c1b6ea057a6d5394161af7a8a6325fb6794ce.png)
Finally,![\[\sum_{k=1}^na_k\le3\cdot\frac1b\sum_{k=1}^nka_k\implies\left(\sum_{k=1}^na_k\right)^3\le\frac94\left(\sum_{k=1}^nka_k\right)\left(\sum_{k=1}^na_k^2\right).\]](http://latex.artofproblemsolving.com/a/4/a/a4a7d4d895bb0d3dbcda0e00aaeb608c648afb0b.png)
be non-negative real numbers such that![\[2a+2b+2c+2d+ab+bc+cd+da+3=abcd.\]](http://latex.artofproblemsolving.com/d/e/8/de8ccd816cfe032a97d940982192b1abea21dc18.png)
Prove that![\[\sqrt[4]{abc}+\sqrt[4]{bcd}+\sqrt[4]{cda}+\sqrt[4]{dab}\le\sqrt[4]{27(1+a)(1+b)(1+c)(1+d)}.\]](http://latex.artofproblemsolving.com/a/f/c/afc556cd8f5c7f4dedf153ffb92fb1f978c56823.png)