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.
be a scalene triangle with circumcenter 
, incenter 
, and incircle 
. Let touch the sides 
, 
, and 
at points 
, 
, and 
respectively. Let 
be the projection of to 
. The line 
intersects the circumcircle of again at point 
. The circumcircles of 
and 
intersect again at points 
and 
respectively. Prove that the lines 
, 
, and 
are concurrent.
be positive real numbers such that : 
. Prove that : 
and 
such that divides 
. Show that there exists integers 
, 
, 
satisfying 
.
be an acute triangle with 
. Let 
be the centroid of triangle 
and let be the foot of the perpendicular from 
to side 
. The median 
intersects the circumcircle 
of triangle at a second point 
. Let 
be the point where intersects . The line 
intersects the circle at a point 
, such that lies between and . Prove that the points 
and lie on a circle.
be non-negative real numbers such that 
Prove that
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 
.
be the set of positive real numbers. Determine all functions 
such that ![\[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]](http://latex.artofproblemsolving.com/e/7/1/e71b6e0d0b858eb46b81149f1e6be8c41e13d301.png)
for every 
.
, and be pairwise distinct nonnegative real numbers. Prove that![\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\]](http://latex.artofproblemsolving.com/2/0/9/209f9807481e7a1e97f023bf65a603cb462c62cb.png)
(Karl Czakler)
that can be written in the form![\[
n = \frac{a^2 - b^2}{b},
\]](http://latex.artofproblemsolving.com/f/9/0/f90a3ce2e1b51d6e1cb53a15cad6f36c2b30492a.png)
where and are positive integers.. Initially, the numbers 
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
, or 
as 
, or 
as 
.) is it possible to end with only one number remaining on the board?
st sees the other three, the 
nd sees the rd and 
th, and the rd sees only the th. The devil has 
hats, numbered from to , he puts one hat on each wise man, and hides the extra 
hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
, and are such that 
. Prove that ![\[\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}.\]](http://latex.artofproblemsolving.com/3/e/4/3e4b336137b17e55bf172d45bd18fe6d5561a827.png)
be real numbers such that 
.![\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]](http://latex.artofproblemsolving.com/5/a/8/5a83eaa7470bcca6546599ff4e296e4efec69b16.png)
and determine all the cases when the equality occurs.
, let be the midpoint of side 
. Let the line perpendicular to at point intersect 
at 
. If 
is tangent to at 
, find 
.
.
for 
means that 
.
. Then, we wish to prove that 
. By AM-GM, 
so it suffices to prove that 
, which is true because 
.
, we just need to prove that 
implies that 
. Expanding gives 
as desired.
, the assertion is true. We will now prove that the assertion is true for .
such that 
, 
, 
don't form the side lengths of a triangle, then 
and 
by AM-GM so 
as desired.
. By AM-GM and our claim, 
as desired.
are equal to , , 
where , are positive reals and 
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.
. Then ![\[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]](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 
, a contradiction. 
and 
? I originally tried using 
's, which was very close but broke slightly for 
. But if we use arbitrary coefficients and 
, it suffices to maximize the function ![\[f(x) = \sqrt x + 2\sqrt{2 - x},\]](http://latex.artofproblemsolving.com/c/3/7/c37b71145f5b1d1c15cfcfa44cbe556c1213ebbe.png)
which by C-S or similar methods is greatest when 
. This fits perfectly when .
. Assume to the contrary that there exists some tuple 
such that the inequality is satisfied and 
. We show that if 
we can form a similar tuple with 
elements. Of course, just remove 
, the decrease on the left is 
, the decrease on the right is 
by AM-GM, so the inequality is preserved. Now we can just reduce to the case, where we MUST have 
. Since 
are increasing for 
, if the inequality holds true for 
, it must hold for 
. Now plugging this value of gives 
, which is NEVER true by AM-GM, so the original inequality can never hold in the conditions described.
WLOG we can assume is the largest. Thus we need to show 
. If we rewrite as a quadratic function in respect to (considering and as parameters) we have:
We however know:
.WLOG we can assume 
. We know that 
. We will first show that 
. This follows from the fact:
From this it directly follows that 
, implying that 
are the sides of a triangle for 
. We are now left to show that 
from which the desired will follow (because 
are the smallest). For simplicity let 
. FTSOC let 
for some 
. We know that by Cauchy-Schwarz 
. We will now show that 
.
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 
, 
and that 
we get a contradiction,
that 
, thus I will
, we get that proving that is equivalent to proving![\[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]](http://latex.artofproblemsolving.com/d/6/3/d6347eedbee72e919170ae49d15d63dc6872409c.png)
When and is a positive real. Thus the above is equivalent to:![\[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]](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 
, thus its equivalent to:![\[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]](http://latex.artofproblemsolving.com/3/d/9/3d9bcb384a9e7159c50d6748733c5ecf1e01c2f4.png)
Which is clearly true.
, and FTSOC let 
. The condition then gives by Cauchy that 
where the summation is taken over all ordered pairs in . Now notice that 
has no integer solutions for 
, so we just need to prove that 
by AMGM. Also, 
by Cauchy so this reduces into 
which is obvious.
that if there exists at least one triple 
that are not the sides of a triangle, then 
For 
we must show that if 
cannot be the sides of a triangle, then 
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 which are not the sides of a triangle, and 
But by our induction above, we have that 
a contradiction. Therefore, all are the sides of a triangle. QED
such that 
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 
increases. Thus, we can assume that 
This clearly implies that 
Hence, we are done.
,![\[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\]](http://latex.artofproblemsolving.com/1/a/b/1ab410ebc17d9f6c38a3e7ae0f6efa955ffddfd3.png)
Show that 
.
.
,
,
.
are the three sides of some triangle for all 
.
terms is at least 
,
and the inequality will be violated.
.
when 
,
,
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)
,
.
.
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.
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:
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 ![\[
\begin{array}{c|ccccccc}
& \frac{1}{t_1}
& \frac{1}{t_2}
& \frac{1}{t_3}
& \frac{1}{t_4}
& \frac{1}{t_5}
& \cdots
& \frac{1}{t_n} \\
\hline
t_1 &1 &a &b &c &d & &v \\
t_2 &a' &1 &g &h &i & &w \\
t_3 &b' &g' &1 &l &m & &x \\
t_4 &c' &h' &l' &1 &p & &y \\
t_5 &d' &i' &m' &p' &1 & &z \\
\vdots & & & & & &\ddots & \\
t_n &v' &w' &x' &y' &z' & &1 \\
\end{array}
\]](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 
,