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.
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
be a triangle. Let 
be a points on the segment 
such that 
. Let 
be the mid-point of 
. Let 
intersect 
in 
and 
in 
respectively. Determine 
.
, Its incircle 
and 
escircle 
are tangent to at 
and 
respectively. 
intersects line at 
. Line intersects at , and line intersects at 
. Line 
intersects at 
. Prove that 
.
that satisfy the following condition for any real numbers 
and 
such that:
be a triangle with circumcenter 
. Let 
be the midpoints of 
and 
respectively and let 
be the projection of on 
. Let be the projection of 
on 
. Let 
intersect 
at points 
and 
. Let 
intersct at points 
. Let 
intersect 
at 
and 
intersect at 
. Then show that 
are concurrent.
and 
respectively be the incentre, circumcentre and circumcircle of triangle . Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
be a sequence that consists of an initial block of 
positive distinct integers that then repeat periodically. This means that 
are distinct positive integers and 
for every positive integer . The terms of the sequence are not known and the goal is to find the period . To do this, at each move it possible to reveal the value of a term of the sequence at your choice. is one of the first 
prime numbers, find for which values of there exist a strategy that allows to find revealing at most 
terms of the sequence.
such that, for all real numbers 
with 
, 
of positive integers for which there exist positive integers 
and 
such that
holds for all integers 
(Note that 
denotes the greatest common divisor of integers 
and 
)
and 
equals one, then 
not 
But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
not But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
and be positive numbers . Prove that 
be positive numbers . Prove that 
and be positive numbers . Prove that 
, the Cauchy-Schwarz's Inequality gives that
It is easy to show that 
. Therefore, the inequality ![$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}$](http://latex.artofproblemsolving.com/f/4/1/f41b3d1ee10a3656bbae0ce631a1c38b24d6b96c.png)
implies![\begin{align*}\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{\frac{9}{\sqrt[3]{(abc)^2}}}{3+\frac{3}{\sqrt[3]{abc}}}=\frac{3}{\sqrt[3]{abc}(\sqrt[3]{abc}+1)}\ge\frac{3}{1+abc},
\end{align*}](http://latex.artofproblemsolving.com/8/4/b/84b96b1dabde2948d23d6ab6fcd9e2ac53f94973.png)
where the last inequality holds since 
. We are done. 
. Prove that
(Lijvzhi)
Prove that![\[ \frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}\geq \frac 4{ 2+abcd}\]](http://latex.artofproblemsolving.com/0/4/d/04d223126d59422752a97e762b601cc298191683.png)
Solution of ytChen 
. Using the Cauchy-Schwarz's Inequality and AM-GM Inequality, we obtain![\begin{align*}\frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}=&\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{z^2}{2z+1}+\frac{u^2}{2u+1}\\
\ge&\frac{(x+y+z+u)^2}{2(x+y+z+u)+4}\\
=&\frac{1}{\frac{2}{x+y+z+u}+\frac{4}{(x+y+z+u)^2}}\\
\ge&\frac{1}{\frac{1}{2\sqrt[4]{xyzu}}+\frac{1}{4\sqrt[4]{(xyzu)^2}}}\\
=&\frac{4}{2\sqrt[4]{abcd}+\sqrt[4]{(abcd)^2}}\\
\ge&\frac{4}{2+abcd},
\end{align*}](http://latex.artofproblemsolving.com/3/e/5/3e5b9b4d313d736f137b85c77060bd4cc4455000.png)
where the last inequality holds since ![$w^4+2-w^2-2w=(w-1)^2(w^2+2w+2)\ge0\ \left(\forall w=\sqrt[4]{abcd}\right)$](http://latex.artofproblemsolving.com/d/3/b/d3b87e901d52d2a6b3ba7dd5104500674c68b4ed.png)
. We are done.
. Prove that(Lijvzhi)
. Using the C-S and AM-GM , we obtain![\begin{align*}\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\\=&\frac{b_1^2}{(n-2)b_1+1}+\frac{b_2^2}{(n-2)b_2+1}+\cdots+\frac{b_n^2}{(n-2)b_n+1}\\
\ge&\frac{(b_1+b_2+\cdots+b_n)^2}{(n-2)(b_1+b_2+\cdots+b_n)+n}\\
=&\frac{1}{\frac{n-2}{b_1+b_2+\cdots+b_n}+\frac{n}{(b_1+b_2+\cdots+b_n)^2}}\\
\ge&\frac{1}{\frac{n-2}{n\sqrt[n]{b_1b_2\cdots b_n}}+\frac{1}{n\sqrt[n]{(b_1b_2\cdots b_n)^2}}}\\
=&\frac{n}{(n-2)\sqrt[n]{a_1a_2\cdots a_n}+\sqrt[n]{(a_1a_2\cdots a_n)^2}}\\
\ge&\frac{n}{n-2+a_1a_2\cdots a_n},
\end{align*}](http://latex.artofproblemsolving.com/8/1/2/81201ba705423fef3914cd56b28fac342bfc815a.png)
where the last inequality holds since 
![$\forall w=\sqrt[n]{a_1a_2\cdots a_n}$](http://latex.artofproblemsolving.com/a/b/f/abf9b837da0fb966837e253cac8a8ebe7fc0e365.png)
. We are done. 
Prove that
and let 
and 
and 
respectively. Let the circle with centre 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
s
. Prove that
Let 
. Prove that
. Prove that