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 
and 
.
satisfies the following property:![\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]](http://latex.artofproblemsolving.com/3/b/d/3bd9fbf82b0b1d98a3c68745bb3f159cd9e2529a.png)
and for all 
.
be a real numbers such that 
Then find maximum value of 
segments 
and 
are parallel. Angle bisectors of 
and 
meet at 
. Angle bisectors of 
and 
meet at 
. Prove that 
is cyclic
such that 
is even , 
is odd and 
is divisible by 
.Prove that there exists a prime number 
such that is divisible by 
and 
, then find the maximum value of ![$\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$](http://latex.artofproblemsolving.com/b/f/a/bfa17e8bb4cc05a84ed6a757f506d6d0ed2b0081.png)
such that there exists a prime number , such that 
is a power of 
. is a number of the form 
where is a positive integer.
be an acute triangle with orthocenter 
and circumcenter 
. Let 
and 
be points on the line segments 
and respectively such that 
is a parallelogram. Prove that 
.
, let 
be the midpoint of side 
. Let the line perpendicular to at point 
intersect 
at 
. If 
is tangent to at 
, find 
.
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line. and are positive integers, we say that almost divides if divides at least one of 
and 
. We call a positive integer almost prime if the following holds: for any positive integers 
such that almost divides 
, we have that almost divides at least one of and . Determine all almost prime numbers.
is an odd number, let 
be a positive integer. If 
then 
such that 
. Find the minimum: 
be an integer. Let 
, 
, ..., 
be positive real numbers such that ![\[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\]](http://latex.artofproblemsolving.com/1/a/b/1ab410ebc17d9f6c38a3e7ae0f6efa955ffddfd3.png)
Show that 
, 
, 
are side lengths of a triangle for all 
, 
, 
with 
.
. it is easy to see that the right is strictly increasing in 
. so if we show that the reverse inequality holds for 
, we are done. a quick calculation is enough to show that(it is too boring to post). but if we know is true for n-1, then let's do the following computation:
,
be an integer. Suppose the inequality
. Prove that 
are the three sides of some triangle for all 
.
terms is at least 
, the whole sum will be at least 
and the inequality will be violated.
. Then 
when 
, we will be done by the argument above.
, so 
and 
verify 
then any three of them are sides of a triangle. I think I've managed to solve this problem and if my solution is correct, then it's a really hard and nice problem.
![\[ (t_1+ \cdots + t_n) ( \frac{1}{t_1} + \cdots + \frac{1}{t_n}) = n^2 + \sum_{i<j} (\sqrt{\frac{t_i}{t_j}} - \sqrt{\frac{t_j}{t_i}} )^2 \]](http://latex.artofproblemsolving.com/8/2/e/82e1a00585a2b5adbd6c59119f4bf55c41b04f5e.png)
then![\[{{ (\sqrt{a/b} - \sqrt{b/a})^2 + (\sqrt{a/c} - \sqrt{c/a})^2 =
\frac{ (a-b)^2}{ab} } + \frac{ (a-c)^2}{ac} }
\geq \frac{(a-b)c}{ab} + \frac{(a-c)b}{ac}
= c/b + b/c -(b+c)/a \geq 2 - 1 = 1 \]](http://latex.artofproblemsolving.com/3/c/1/3c1c7545ca6ebf41b70af49f67869215ec2039d5.png)
(t_1+t_2+...+t_n)(1/t_1+1/t_2+...+1/t_n)=(3\times(t_1+t_2+t_3)/3 +t_4+...+t_n)(3\times(1/t_1+1/t_2+1/t_3)/3 +1/t_4+...+1/t_n)\geq(3\times \sqrt((t_1+t_2+t_3)/3\times(1/t_1+1/t_2+1/t_3)/3) +1+...+1)^2
and 
Prove that
.
.n^2+1>(\sqrt((t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)) +n-3)^2(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<(\sqrt(n^2+1) -(n-3))^2n\geq3, it follows(\sqrt(n^2+1) -(n-3))^2<10 <=> 2n^2 -6n<(2n-6) \sqrt(n^2+1)
(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<10
t_1 \geq t_2 \geq t_3. Then
t_1+t_2 \geq t_3, t_1+t_3 \geq t_2
t_1>t_2+t_3. Then
(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)-((t_2+t_3)+t_2+t_3)(1/(t_2+t_3) +1/t_2+1/t_3) =(t_1-t_2-t_3)(1/t_2+1/t_3-1/t_1) \geq 010>(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3) \geq 2(t_2+t_3)(1/(t_2+t_3)+1/t_2+1/t_3) <=> 4>(t_2+t_3)(1/t_2+1/t_3) \geq 4
x_1,x_2,...,x_nverify
k(n)>(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)then any three of them are sides of a triangle.
(n+\sqrt{10}-3)^2. I think this one is much more interesting.
f(x_1,x_2,...,x_n)=(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)where
x_1<=x_2<=...<=x_n.
k(n)=(n+\sqrt{10}-3)^2might be found elementary. From the solution of Sung-yoon Kim:
n^2+1>(\sqrt((x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3))+(n-3)^2(x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3)<(\sqrt(n^2+1) -(n-3))^2(n^2+1), use that
k(n)and instead of
x_3, use
x_n:
(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2x_1+x_2<=x_nthen:
(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)>=10, so we must have
10>=(\sqrt(k(n)) -(n-3))^2,(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2, then:x_1+x_2>x_n, so for any
1<= i, j, k <= n:
x_i+x_j>x_k.
k(n)=min f(x_1,x_2,...,x_n),
x_1<=x_2<=...<=x_nand
x_1+x_2<=x_n, we may get:
k(n)=f(1,1,2\sqrt(2/5),...,2\sqrt(2/5),2), where
2\sqrt(2/5) appears (n-3)times.
Medges ,
M>=3so if the same inequality holds for any
n>=Mpositive reals (of course with something like
k(n,M)in the RHS) then any
Mfrom our
nreal numbers can be the side lengths of a polygon with
Medges. Solution is basically the same.
.
,
.
.
do not form the sides of a triangle. Then
.
for 
.
.
.
so it suffices to prove that 
,
.
implies that 
.
as desired.
,
such that 
and 
by AM-GM so 
as desired.
.
as desired.
where 
is nonnegative. By Cauchy-Schwarz, we have
Now
by AM-GM. To reach contradiction it is now sufficient to show 
which is true. We are done.
.
Thus, we want
if 
Since 
,
which is clearly a contradiction. Thus, for all 
,
and 
are side lengths of a triangle which implies that 
,
and 
are as well. Swapping the indices we get the conclusion for all 
,
,
and 
.
.
and do a second degree equation in 
.
,
so 
which is a contradiction from Cauchy.
.![\[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]](http://latex.artofproblemsolving.com/0/a/c/0acd6d3ef414f23102f5794365916ac04db3c1c9.png)
and ![\[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \geq \frac{1}{t_1} + \frac{4}{t_2 + t_3}\]](http://latex.artofproblemsolving.com/a/7/2/a7235fbb9869141c281b54ebbd1937a89bad445e.png)
by Cauchy-Schwarz, so we have ![\[(t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\]](http://latex.artofproblemsolving.com/7/6/6/76643350b294b5322309ee8358dca57f236afcb5.png)
![\[ \geq \left(\frac25 t_1 + \frac85(t_2 + t_3) + t_4 + \dots + t_n\right)\left(\frac{1}{t_1} + \frac{4}{t_2 + t_3} + \frac{1}{t_4} + \dots + \frac{1}{t_n}\right)\]](http://latex.artofproblemsolving.com/5/6/b/56be74ad5093a6bdb0b8c517fedd46f1846faee2.png)
![\[ \overset{\text{C-S}}{\geq} \left(\sqrt{\frac25} + \sqrt{\frac{32}{5}} + (n - 3)\right)^2 = \left(n + \sqrt{10} - 3\right)^2.\]](http://latex.artofproblemsolving.com/7/c/0/7c0c0364f68030796905b430e213165b3fd94381.png)
But 
is strictly increasing, so 
,
and 
?
'
.
,![\[f(x) = \sqrt x + 2\sqrt{2 - x},\]](http://latex.artofproblemsolving.com/c/3/7/c37b71145f5b1d1c15cfcfa44cbe556c1213ebbe.png)
which by C-S or similar methods is greatest when 
.
such that the inequality is satisfied and 
.
elements. Of course, just remove 
,
by AM-GM, so the inequality is preserved. Now we can just reduce to the 
.
are increasing for 
,
,
.
,
WLOG we can assume 
.
as parameters) we have:
We however know:
.
.
.
.
From this it directly follows that 
,
are the sides of a triangle for 
.
from which the desired will follow (because 
are the smallest). For simplicity let 
.
for some 
.
.
.
From here it follows that:
![\[\left(t_1 + t_2 + \dots + t_n\right)
\left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) = n+\sum_{i, j \in \{1, 2, \dots, n\}, i\neq j} \frac{t_i}{t_j} +\frac{t_j}{t_i}\]](http://latex.artofproblemsolving.com/6/0/3/603a241e4653fd3df03efa30e822979ccf517f55.png)
Thus we get if for any 
we get a contradiction,
that 
,
,![\[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]](http://latex.artofproblemsolving.com/d/6/3/d6347eedbee72e919170ae49d15d63dc6872409c.png)
When ![\[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]](http://latex.artofproblemsolving.com/e/6/e/e6e5ffd6f245a148efe013b7f83b0e5e68252740.png)
![\[\frac{b+a}{ab}-\frac{1}{a+b}\]](http://latex.artofproblemsolving.com/4/9/7/497c7d2bc149121cfb147ceba6006cffeb0b1315.png)
The above is minimised when 
,![\[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]](http://latex.artofproblemsolving.com/3/d/9/3d9bcb384a9e7159c50d6748733c5ecf1e01c2f4.png)
Which is clearly true.
,
.
where the summation is taken over all ordered pairs in 
has no integer solutions for 
by AMGM. Also, 
by Cauchy so this reduces into 
which is obvious.
that are not the sides of a triangle, then 
For 
we must show that if 
WLOG, assume that 
then we have that 
Also, because our inequality is homogenous, assume that 
Letting 
it suffices to show that 
But by AM-GM, 
so it suffices to show that 
But, 
by 
applications of AM-GM at the end. Therefore, our base case is proven.
Now, assume that for 
our proposition holds. Consider the sequence 
of positive real numbers. Then if there is at least one triple, say WLOG 
which are not the sides of a triangle then we have by our inductive hypothesis that
Therefore, our induction is complete. QED
But by our induction above, we have that 
a contradiction. Therefore, all 
Mainly for storage.
.
.
,
and 
.
.
.![$(s+u+v)(\frac{1}{s} + \frac{1}{u} + \frac{1}{v}) \ge (s+u+v)(\frac{4}{s+u} + \frac{1}{v}) = 5 + \frac{4v}{s+u} + \frac{s+u}{v} \ge 5 + 5 \cdot \sqrt[5]{\frac{v^3}{(s+u)^3}} \ge 10$](http://latex.artofproblemsolving.com/3/2/2/322de8dea08b6057f70ea16e1396449a2cc2af43.png)
when 
.
We rephrase the problem as such:
be a table with 
the number 
is written. A table is "good" if all of the numbers in it sum up to a number strictly less than 
.
for all 
.
written term by term:![\[
\begin{array}{c|ccccccc}
& \frac{1}{t_1}
& \frac{1}{t_2}
& \frac{1}{t_3}
& \frac{1}{t_4}
& \frac{1}{t_5}
& \cdots
& \frac{1}{t_n} \\
\hline
t_1 & & & & & & & \\
t_2 & & & & & & & \\
t_3 & & & & & & & \\
t_4 & & & & & & & \\
t_5 & & & & & & & \\
\vdots & & & & & & & \\
t_n & & & & & & & \\
\end{array}
\]](http://latex.artofproblemsolving.com/2/2/c/22c8c239fc3d7d4edd456b3ca64c7d274d2fa718.png)
Let us now start filling in the table with numbers - we immediately notice that the cells along the main diagonal all have the number 
inside of them, so we can immediately fill these ones up:![\[
\begin{array}{c|ccccccc}
& \frac{1}{t_1}
& \frac{1}{t_2}
& \frac{1}{t_3}
& \frac{1}{t_4}
& \frac{1}{t_5}
& \cdots
& \frac{1}{t_n} \\
\hline
t_1 &1 &a &b &c &d & &v \\
t_2 &a' &1 &g &h &i & &w \\
t_3 &b' &g' &1 &l &m & &x \\
t_4 &c' &h' &l' &1 &p & &y \\
t_5 &d' &i' &m' &p' &1 & &z \\
\vdots & & & & & &\ddots & \\
t_n &v' &w' &x' &y' &z' & &1 \\
\end{array}
\]](http://latex.artofproblemsolving.com/b/a/0/ba08e85c9ac36fc7a91ac7b4c59b0528aa25db6f.png)
Now we observe the following
etc.)
for all real 
.
with cell 
)
,
cells (two by two symmetric in regards to the main diagonal) with a sum of the numbers in them at least 
such that 
- assuming that such exists would invariably lead to a contradiction, thus completing the problem.
.
(note that if there exists a triple that does not form a triangle, then the triple 
also does not form one) and we have thus reduced the problem to proving that
Since the expression on the left is scalable, we can assume that 
,
and 
.
From here, we use the useful lemma 
(can be derived - pun intended - from the second derivative of the LHS function and proving that it is reached for 
) to get the inequality we need to prove to be simply
Since this function is increasing for 
(this can be verified by expressing the difference 
)
at 
,
cells with a sum of their numbers at least 
,
Then, we have that:
Thus, all that is left is to prove that 
Expanding, we have that 
Now, differentiate w.r.t. 
We get 
We claim that 
However, this is clearly true because 
while 
Thus, 
is increasing as 
This clearly implies that 
Hence, we are done.