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 non-negative real numbers satisfying 
and 
. Prove that:![\[
3(ab + bc + ca) \left( \frac{a + b}{b + c} + \frac{b + c}{c + a} + \frac{c + a}{a + b} \right) \geq 23 + \frac{t^3(21 - 5t)}{2(3 - t)}
\]](http://latex.artofproblemsolving.com/3/b/8/3b8980f274c65a1370a99beb422369d5e4352ca2.png)
where ![\( t = \sqrt[3]{abc} \)](http://latex.artofproblemsolving.com/b/5/c/b5c0479cd48ec8d01ce071924b1e388c84b413fa.png)
.
are real numbers such that
,
.
.
, 
and 
are three real numbers, all different from 
, such that 
, then prove that 
. (With the 
sign for cyclic summation, this inequality could be rewritten as 
.), and .
with 
the following inequality holds:
, 
, 
, the inequality![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](http://latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
, 
, and 
with 
prove that
be the circumcenter and 
the orthocenter of an acute triangle 
Prove that the area of one of the triangles 
and 
is equal to the sum of the areas of the other two.
with any line through the centroid 
and I have no clue as to why that is true. Plz help
such that for all 
,![\[xf(x+f(y))=(y-x)f(f(x)).\]](http://latex.artofproblemsolving.com/e/4/a/e4a3bbb8b91d2aa62d699c24df342fa59be71915.png)
board, there is initially a tree of height 
. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
. Determine the largest 
such that the gardener can ensure there are eventually majestic trees on the board, no matter how the lumberjack plays.
. Let 
be its incenter, 
be its 
-excenter, 
be midpoint of arc 
, 
is the midpoint of 
.
be the intersection of the line 
with 
, 
be the intersection of the angle bisector of 
with the line .
,the intersection of line 
with 
and 
,the intersection of 
and 
such that 
. Ana wins if the points 
are collinear. Who has a winning strategy?
be positive real numbers so that 
. Prove that![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
be a scalene triangle with incircle 
. Denote by 
the midpoint of arc in the circumcircle of , and by 
the point where the -excircle touches . Suppose the circumcircle of 
meets again at 
and intersects at two points 
, 
.
or 
is tangent to .
, let 
be a point on the segment and let be the intersection point of 
with the diagonal 
. The line perpendicular to the segment through intersects the side 
at point 
. Let be a point on the segment 
such that 
. Let be the intersection point of the diagonal 
and the segment 
.
and 
are concyclic.
that satisfy the following condition: for all and relatively prime to 
we have ![\[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]](http://latex.artofproblemsolving.com/e/8/5/e85c5b249ac8e75b45a525b8786414798a4921fe.png)
be the set of the positive real numbers. Find all functions 
such that 
for all positive real numbers and .
![\[
f(x)\geq x \text{ }(1).
\]](http://latex.artofproblemsolving.com/b/a/1/ba142248678f97e011c46d9973d172fbaeba6a6e.png)
, took any 
. Then, for 
![\[
f(u) = f(f(x)f(b))+x > x
\]](http://latex.artofproblemsolving.com/1/b/b/1bb5fb34ed534c417a4b47d986df8b7f3a934a86.png)
![\[
\Rightarrow f(u) > x, \forall x \in \mathbb{R}_{>0}
\]](http://latex.artofproblemsolving.com/e/4/d/e4df1325bb8621e167a47530850ecbfa53c39476.png)
, then we can substitute 
in the previous inequality and get 
, which contradicts . This implies on the original equation![\[
f(xy+f(x) \geq f(x)f(y) + x \text{ }(2)
\]](http://latex.artofproblemsolving.com/9/5/3/9536094eee98182b2efce0905eed95bc6490c347.png)
.
, took 
in 
![\[
f(b) \geq f(a)f(b) + a \Rightarrow f(a) \leq 1 - \frac{a}{f(b)} < 1.
\]](http://latex.artofproblemsolving.com/3/e/7/3e78fec32a7245cf96e1a6d7b2183888146fe03c.png)
, if 
, then 
. Took inequality in the form![\[
f(x) \leq \frac{f(xy+f(x))-x}{f(y)} \leq \frac{f(xy+f(x))-x}{y}
\]](http://latex.artofproblemsolving.com/f/b/f/fbf9d4ca7e3a47e3f573d85818e9c4ddb23f8a1e.png)
, took any such that 
. Then![\[
f(x) < \frac{1-x}{y} \Rightarrow y < \frac{1-x}{f(x)}.
\]](http://latex.artofproblemsolving.com/8/5/f/85f14538e744499ef94190f6fe1995b77f331148.png)
, we have 
. Which means that doesn't exists such that 
. This implies![\[
\frac{1-x}{f(x)} \geq \frac{1-f(x)}{x}
\]](http://latex.artofproblemsolving.com/e/b/d/ebd0f218173ac1bf5ae5c43c918336ed70c93943.png)
![\[
(f(x)-x)(f(x)+x-1)\geq 0
\]](http://latex.artofproblemsolving.com/b/0/8/b083a5cb7438dfe64bfb91e7a91b5ed8ef142307.png)
. Now we'll prove that for every 
. Suppose for sake of contradiction that it's false. Then took 
such that 
. This implies 
. Take 
in . As 
, we have 
. Then![\[
1>f((2/3)u+f(2/3))\geq f(2/3)f(u)+2/3
\]](http://latex.artofproblemsolving.com/4/0/a/40a9e5ef6b6b19bc106d539c4920358dc7766b7f.png)
![\[
1/3 > (1-u)f(2/3)\geq f(2/3)(1/2) \Rightarrow f(2/3)<2/3
\]](http://latex.artofproblemsolving.com/c/8/0/c804a5058a6e2dce13128f1cf9ad63b20c2b9524.png)
. Now we can finally prove that the only solution is 
. First, take fixed points and 
. The original equation for these values give us![\[
f(x(y+1))=f(xy)+x=x(y+1)
\]](http://latex.artofproblemsolving.com/7/8/7/787d4d31d34ee5d296f77e26aaf9e9cc342d7c2d.png)
is also a fixed point. Suppose that exists a maximal such that for every 
, - the last fact showed that exists such that makes this interval non-trivial. But observe that for every 
- which is a non empty interval - we also have 
. First, 
- just took 
sufficiently big such that 
and then took 
and 
. Now, took 
and 
- which satisfies 
- in the previous result and we have . So, we got an interval 
such that every point inside it is fixed, and it's stricly bigger then the previous one, which contradicts the hypothesis of maximal. So, as such doesn't exists, we have that every point in 
is fixed.
?
, which is trivially a solution.
, we know from 
that 
, which is clearly absurd.
for some 
, as otherwise we arrive at our solution 
, we have by 
that:
and 
so that 
, so that 
We have 
If 
, we're done. Else we have 
But then we have 
so we're again done.
for sufficiently large 
be a real number so that 
, which exists by Lemma 
We have that 
. For sufficiently large, we have 
and so the lemma is proven.
so that for all 
, we have 
Fix a constant 
, there exists some 
with 
By 
, we have that 
, so we just need to show that 
i.e. 
We have that
Let 
. Then, by the Claim applied 
times, we know that there is some real number 
so that 
. However this is negative, absurd. for some 
, and so the only solution is .
...
?. For sufficiently large, we have and so the lemma is proven.
?
?
so that 
. If 
, that means that 
, so 
?, we have that 
Since 
for all 
, this means that 
for sufficiently large 
.
cannot hold for any because in that case we would be able to produce tending to 
.
denote the assertion 
. First, we give a nontrivial bound of .
for any .
, then 
gives 
which is contradiction. 
for some 
. Call a positive real tasty if and only if there exists real which 
for any 
. We prove two claims about tasty numbers which eventually lead us to deduce that grows faster than linear.
is tasty.
, the claim follows from
is tasty, then any constant 
is tasty.
, then
By varying , we see that 
when 
. This gives the result. 
for any . Now choose 
so that 
. This is possible since 
. Then consider
which means 
, contradiction.
for all .
for some . Then 
gives 
or 
, contradiction. 
exists and equals zero.
. Indeed, whenever 
, we can find such that 
; then ![\[ f(y) = f(xy+f(x)) = f(f(x)f(y))+x > f(f(x)f(y)) \ge f(x)f(y) \]](http://latex.artofproblemsolving.com/2/8/9/28957c8b3a5c9981d7e754ed38349017956b7129.png)
which implies the boxed lemma.
be given and choose 
. Then let 
be any number with 
. Then 
gives ![\[ 1 \ge f(\underbrace{f(x) + \delta \cdot x}_{<1}) = x + f(f(x)f(\delta)) > (1-\varepsilon) + f(x)f(\delta). \]](http://latex.artofproblemsolving.com/1/3/e/13e453cc0aa4e5372e3dd45444224d80dad9aacd.png)
Since 
, we have 
. In conclusion we have prove the statement ![\[ f(\delta) < 2\varepsilon \qquad \forall 0 < \varepsilon < \frac{1}{2} \text{ and } \delta < 1-f(1-\varepsilon) \]](http://latex.artofproblemsolving.com/f/8/e/f8e78ed6ee9efb38ef61a7cb58f8b39ee35dd350.png)
So this implies the limit condition. and consider small ; we have ![\[ x \le f(x) < xy + f(x) \le f(xy+f(x)) = x + f(f(x)f(y)). \]](http://latex.artofproblemsolving.com/d/b/5/db5e73afd97d0e724d2d004a8ad0e8508d870e1a.png)
Since 
by applying the last claim twice, this implies 
for all .
is the only solution, firstly note that it satisfies given equation . be any function satisfying the assertion which we shall call .
then 
.
and hence it is a bijection .
be the set of the positive real numbers. Find all functions such that for all positive real numbers and .
.
so that 
. Then, 
gives 
, a contradiction.
for all 
., choose 
so that 
. Then, ![\[f(y)=f(xy+f(x))=f(f(x)f(y))+x>f(x)f(y)+x>f(x)f(y)\]](http://latex.artofproblemsolving.com/5/d/e/5de748e146e9c682a938ddd87b93890b0bb77d02.png)
and so we are done.
.
arbitrarily small. Then 
. Also choose 
such that ![\[(1-\epsilon)\lambda+f(1-\epsilon)<1\]](http://latex.artofproblemsolving.com/d/d/b/ddb44cc8b9c0d4bf8a163cd31388875534967ed5.png)
Then, 
and 
gives ![\[1>f(x\lambda+f(x))=f(f(x)f(\lambda))+x=f(f(x)f(\lambda))+1-\epsilon\]](http://latex.artofproblemsolving.com/4/b/9/4b9e90c4e550860c0d2f8e2910eb45003d1786a5.png)
so ![\[0<f(f(x)f(\lambda))<\epsilon\]](http://latex.artofproblemsolving.com/c/3/f/c3f7a9b93f08e129b5b4974f6026bd9ec49a3e62.png)
and since 
is arbitrarily small, we have that 
tends to 
. Now, 
gives ![\[x\leq f(x)<x\lambda+f(x)\leq f(x\lambda+f(x))=f((x)f(\lambda))+x\]](http://latex.artofproblemsolving.com/4/d/8/4d8faac9980d6c7f02324a8e75803476c8b35ce3.png)
and since tends to , 
. Switching 
and gives 
. Therefore, we have that 
(since for ) which tends to , if we choose sufficiently small. Therefore, as 
tends to , it's image also tends to , as desired.
and choose 
sufficiently small, then ![\[x\leq f(x)<xy+f(x)\leq f(xy+f(x))=x+f(f(x)f(y))\]](http://latex.artofproblemsolving.com/6/7/b/67b226f57e7d37c3a8499d18b43a020c5db87f77.png)
Also we have 
so 
and so (since the first inequality is actually an equality)
, which clearly works denote the assertion. We present a series of claims to solve this problem... Then 
yields : contradiction.
for all ., consider 
, so 
. Then gives
Thus , since 
.
.
such that there exist arbitrarily small with 
. Let 
and pick a small as described such that 
as well. Then from 
we have the inequality chain
which is absurd. Thus such a does not exist, so we have 
as desired.
. and send 
. Then 
as well, so 
. This means that
On the other hand, we have 
, so if then the above equation cannot be true. for all , as desired. 
., then 
, contradiction. then .
. If we can take 
to force , which forces . is unbounded below.
and such that (possible because ), we get 
.
, we have 
. Let 
, so we have 
.
, 
. It suffices to find such that . Since and 
, this is possible.
for all 
., we have 
. By Claim 2, cannot be less than 1, so 
. Now 
. By induction, .
.
for 
. Then taking 
where 
, we have 
. Taking sufficiently large leads to a contradiction.
, which works.
for all 
. Indeed, if for a , then 
yields , absurd. for all . Indeed, if then 
yields (using Step 1) 
that is as desired.
such that 
for all 
. For this step, take a and 
. By Steps 1 and 2, is positive, and so
and so 
is bounded for all , as wanted. for all . Indeed, fix and take to be pretty small. Then,
when gets closer to , and we are done. is identity everywhere. Since , put 
to infer that 
, and so 
as wanted.
, easy to check. Let 
be the assertion above. for all . Assuming , take 
, we obviously get . such that 
, note taking 
we have![\[f(cy+c+\epsilon) = f((c+\epsilon)f(y)) + c\]](http://latex.artofproblemsolving.com/1/f/9/1f9f7cebcd6b6b91b065b27010b49651689b6b7a.png)
Now taking for arbitrarily large greater than some bound, note we have 
. Thus we can define a sequence 
, such that![\[a_i = \begin{cases}
cy+c+\epsilon, \quad i = 1\\
(c+\epsilon)f\left(\frac{a_{i-1} - c - \epsilon}{c}\right), \quad i > 1
\end{cases}
\]](http://latex.artofproblemsolving.com/6/d/6/6d66c00ad574cbec1d73e6fe67b1871357804a87.png)
This gives 
. Note is strictly increasing and 
is strictly decreasing by a constant, thus a contradiction to 
, i.e .
, easy to check. Let be the assertion above. for all . Assuming , take , we obviously get since 
. such that , note taking we haveNow taking for arbitrarily large greater than some bound, note we have . Thus we can define a sequence , such thatThis gives . Note is strictly increasing and is strictly decreasing by a constant, thus a contradiction to , i.e .
then, 
, 
(By Claim 1,2,3)
is independent of 
is the identity, which works. then choose such that 
, which implies , contradiction. Thus for all .. Suppose . Then choose such that to get .
implies 
for some . If 
then choose large enough such that 
for some 
, which is possible since . Now 
, since 
. Thus 
implies 
. By induction, this holds for all .
then 
. Setting 
gives 
, so ![\[(f(x)f(y)+x)-(xy+f(x))=f(x)(f(y)-y)+(f(x)-x)(y-1)<m.\]](http://latex.artofproblemsolving.com/7/f/3/7f385b41185b2b3c9e98fdcd0b35ba773228f22f.png)
The first term is positive and 
so 
. Taking arbitrarily large implies , as desired.
take 
and we get 
contradiction.
then 
and we get 
by our first claim. Thus, since 
we are done.
exists and is 0.
and choose 
Pick 
which has to exist since 
for 
Then, 
tells us 
or 
so our claim is done.
sufficiently small and we get 
and taking the limiting behavior, we get is squished to become 
![\[f(xy+f(x))=f(f(x)f(y))+x\]](http://latex.artofproblemsolving.com/7/e/5/7e559eaa1594b74346a7083ed124574a37def5c9.png)
The only function holding the equation is . Let be the assertion..
, by 
, we have 
which is a contradiction hence .
., we see that implies 
.![\[f(xy+f(x))\geq x+f(x)f(y)\]](http://latex.artofproblemsolving.com/a/3/f/a3f759e2cb6e67384d8285900cd54e3b7b53c040.png)
Let , pick to get 
thus, . such that 
for all 
.
, then pick between them in order to observe ![\[1>f(xy+f(x))\geq x+f(x)f(y)\geq x+yf(x)>x+1-x=1\]](http://latex.artofproblemsolving.com/7/3/9/739c97eecae1d3f4d1ad1da677f11d43df039714.png)
which is impossible. Hence 
for . This implies 
. We either have or 
for all 
. Suppose that 
so 
. We see that 
for 
gives![\[1>f(\frac{2}{3}.\epsilon+f(\frac{2}{3}))\geq \frac{2}{3}+f(\epsilon)f(\frac{2}{3})\geq \frac{2}{3}+\frac{2}{3}(1-\epsilon)=\frac{4}{3}-\frac{2\epsilon}{3}\implies \epsilon\geq \frac{1}{2}\]](http://latex.artofproblemsolving.com/5/8/6/58637608b18611133b214d9535321e8f4fda36ba.png)
Which is not true. Hence 
satisfies the conditions. is injective.
, then 
and 
give 
which contradicts with our assumption. for all 
where 
, then for 
, implies ![\[f(a(b+1))=a+f(ab)=a(b+1)\]](http://latex.artofproblemsolving.com/3/1/1/311eb8ed971f7186e7cb6286c3edd3d200437e47.png)
attains all values in 
. We can extend the bound from 
to 
. In the sequence 
we see that each term is at least 
more than the previous term hence 
term will be at least 
until it reaches to 
. Hence this will be larger than . for all 
, then 
for sufficiently small and 
, we see that 
or 
. Since is injective, 
for , too. for all positive reals as desired.
would give a contradiction.
Take 
, 
holds and consider the sequence
Clearly 
is increasing and 
which is absurd.
is impossible?
we can choose a 
which is clearly impossible .
in claim 1.
such that 
as 
and 
for all 
.
and this along with our choice of 
and hence it is true for all 
This means 
for all 
,
. Substituting 
we have
By claim 1, as 
we have 
.
,
for some 
and 
Now simply let 
to get 
as required .
and hence 
for all 
.
![\[P(x, y)\]](http://latex.artofproblemsolving.com/b/e/6/be6fce191ed5ad2394f3b47289f836e7445df1d5.png)
![\[\therefore f(k+f(x))> x\]](http://latex.artofproblemsolving.com/9/9/9/999fdaa43457d2ec46f69ccd3bed36b4a417132a.png)
For all 
,
.
,
,
and let 
.![\[P(c, b)\]](http://latex.artofproblemsolving.com/1/a/f/1af392f315d4cd91471b35d12b4f42c7afa4f935.png)
![\[f(\frac{cf(c)}{1-c}+f(c))\geq f(c)f(b)+c\]](http://latex.artofproblemsolving.com/3/5/7/3573afbd13fbc10f8f397a3d14713fdc8752c7a8.png)
![\[f(\frac{f(c)}{1-c})\geq f(b)c+c\]](http://latex.artofproblemsolving.com/5/5/1/551fc230779330d910bbb8ca81ea8a1720d74256.png)
![\[f(b)\geq f(b)f(c)+c\]](http://latex.artofproblemsolving.com/7/7/7/7770c29e4daee6047fcde4c5d0bb73a363e0152a.png)
![\[-c\geq f(b)(c-1)\]](http://latex.artofproblemsolving.com/c/f/5/cf55739e2e78685ade6956d742f3d74b345b0c45.png)
![\[\frac{c}{1-c}\geq f(b)\geq b\]](http://latex.artofproblemsolving.com/f/a/6/fa6ea6d1250b5d5e7f6c27d89f9dbe0b2e4ac908.png)
However 
we get:![\[\frac{c}{1-c}=\frac{f(c)}{1-c}\]](http://latex.artofproblemsolving.com/d/d/9/dd9efcc4770f07b7fe483221bd453e616b4cd23a.png)
Thus:![\[f(c)=c\]](http://latex.artofproblemsolving.com/2/2/c/22c64e97fbc79f98f93e7dc31fe1b0279f9a874c.png)
for all 
.
be a value such that 
.
.![\[P(x, x')-P(x', x)\]](http://latex.artofproblemsolving.com/e/5/3/e53dac02bde976ec77642e7bed1822628257b0b1.png)
![\[f(xx'+f(x))-x'(x+1)=x-x'\]](http://latex.artofproblemsolving.com/f/f/f/fffda7f64e5dae04d8dbaaab272e0d77c7b16104.png)
![\[f(xx'+f(x))=xx'+f(x)\]](http://latex.artofproblemsolving.com/8/7/f/87f582f5f93ee7a5bcd7cffa5b777829ac682c26.png)
Thus