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.
+1 w
, 
, 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
+2 w![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
positive integers. Product of any one of them with sum of remaining numbers increased by 
is divisible with sum of all numbers. Prove that sum of squares of all numbers is divisible with sum of all numbers
be a triangle with 
, and let 
be the foot of the altitude from 
. Let 
be a point in the interior of the segment 
. Let 
be the point on the segment 
such that 
. Similarly, let 
be the point on the segment 
such that 
. Let 
be the point of intersection of 
and 
.
. be a scalene triangle with circumcircle 
. Let be the midpoint of 
. A variable point 
is selected in the line segment 
. The circumcircles of triangles 
and 
intersect again at points and 
, respectively. The lines 
and 
intersect (a second time) the circumcircles to triangles and at and 
, respectively. Prove that as varies, the circumcircle of 
passes through a fixed point 
distinct from 
.
, 
, 
, ... be an infinite sequence of real numbers satisfying the equation 
for all 
, where and are two different positive reals., , , ... be bounded?
be two positive sequences defined by 
and![\[ \begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \]](http://latex.artofproblemsolving.com/2/0/1/2012dc6ff3a41478547cea4aa9b6bccbe17a3623.png)
for all 
.
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)
with 
, 
and 
for all positive integers .
where is a positive integer and is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
such that 
divides 
for all positive integers and with 
.
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
, 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 
be positive reals with 
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)
be two triangles and 
and 
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)
,![\[ (x_1 + 2x_2 + \cdots + nx_n)(x_1^2 + x_2^2 + \cdots + x_n^2) \ge c (x_1 + x_2 + \cdots + x_n)^3 \]](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)