AoPS Academy
be an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
with incircle 
. Let 
be the point on such that the circumcircle of 
is tangent to and let the -excircle be tangent to at 
. Prove that the tangent from to and the tangent from to (distinct from ) meet on the line parallel to and passing through .
such that
for all positive real numbers 
.
we get 
therefore if 
and 
then 
, Proving that 
is surjective.
be a number s.t 
, then 
, so 
we get 
, so for 
.
be any number, and 
sufficiently big numbers so 
, then:
and
and 
Then 
and 
then
I will edit the latex later.. (Edit: latex editted), ![$[(z+1)/2]*f(2x)=f(xf(z)+x)$](http://latex.artofproblemsolving.com/6/9/9/699692977b9bd0cc40c107db97b33c4979fac06f.png)
.
, then 
=> 
since 
.
=> ![$[(z+1)/2]f(2x)=f(x(f(z)+1))$](http://latex.artofproblemsolving.com/7/4/4/74497298947a55c3e649089450fd94fe2254ea33.png)
..........(2)![$x=1, [(z+1)/2]f(2)=f(f(z)+1)$](http://latex.artofproblemsolving.com/9/a/b/9abd596f0d8e3148c7dca6b52e06fc8a5043e200.png)
=> f is surjective on 
.
, take any 
, take s.t 
.![$[(z+1)/2]t=[(z+1)/2]f(2a)=f(xf(a)+x)$](http://latex.artofproblemsolving.com/7/a/5/7a5a171d7263c1374be5e492a0646a983fba51e0.png)
=> ![$[(z+1)/2]t$](http://latex.artofproblemsolving.com/8/7/b/87b092579e311d49709dc5bebc34aa38b037789c.png)
is also in 
.
=> ![$[(z+1)/2]t \in Im(f)$](http://latex.artofproblemsolving.com/3/0/1/301cf96ad644bf14fabbffb0aa0e457516b41af3.png)
=> ... => ![$[[(z+1)/2]^k]t \in Im(f)$](http://latex.artofproblemsolving.com/1/8/2/1821925e8f2a0d3c5d4d0832f2270e112b486da3.png)
. then ![$[[(z+1)/2]^k]t$](http://latex.artofproblemsolving.com/3/c/4/3c41606ed3e080dc555a03b22d21e1d2cb7567fe.png)
can be arbitarily small, but it is in .
so that 
.
, then 
=> 
, since 
. must also exist by bijectivity.
, in particular 
..........(4)
and 
at (2).![$f(\frac{2}{[f(z)+1]})=\frac{2}{(z+1)}$](http://latex.artofproblemsolving.com/2/5/2/25259c1b156f0fac808fd081979e116e8ed60a8c.png)
and ![$f(\frac{[f(z)+1]}{2})=\frac{(z+1)}{2}$](http://latex.artofproblemsolving.com/9/4/b/94ba69dac3819887a6911db8021cba1a61969b5c.png)
.
can take any value from 
.
, exist s.t ![$\frac{[f(z)+1]}{2}=w$](http://latex.artofproblemsolving.com/f/e/0/fe0521d2e578f2c7caddc76f623ab998fe532ed0.png)
![$f(1/w)=2/(z+1)=1/[(z+1)/2]=1/f(w)$](http://latex.artofproblemsolving.com/1/8/9/189cfae391d5d029b3b4e59df032f70f7b0e4bb0.png)
for all 
, then we can take 
, then 
, same equation.
for all 
..........(5)
, then 
=> 
.........(6)
we can choose s.t ![$p=[f(z)+1]/2$](http://latex.artofproblemsolving.com/7/f/2/7f206f1f375eda8d3bad63bd6c4ea5064b3a9e42.png)
.![$f(2p)=f(f(z)+1)=[(z+1)/2]f(2)=f(2)[(z+1)/2]=f(2)f([f(z)+1]/2)=f(2)f(p)$](http://latex.artofproblemsolving.com/6/2/8/62898bf477c726bae1ebbb5cddec8dbcdcb3f7ac.png)
.
for all . for all 
..........(7) (By using similar idea like (5).)
=> 
for all 
..........(8)
, we ignore 
from now.![$[f(c)+1]f(a+b)=f(ac+b)+f(bc+a)$](http://latex.artofproblemsolving.com/3/2/b/32b00f10404594c6c71acdbc6fdbb52fdc8765cc.png)
for all 
..........(**)![$a=b=1, [f(c)+1]f(2)=2f(c+1)$](http://latex.artofproblemsolving.com/b/f/e/bfe1ae0b139fbc72ffe050820da9659daa360fe8.png)
..........(9)
, then by (7), 
.
, then 
=> 
.
, then ![$a^2=f(4)=a[a(a+1)/2+1]/2$](http://latex.artofproblemsolving.com/9/8/d/98dd32802b0dfc1a9be7a6e829909783e654fc06.png)
=> ![$2a=[a(a+1)/2]+1$](http://latex.artofproblemsolving.com/7/6/3/7634e7f81fce3a9a54a50c2ed1e1504d2e4951ee.png)
.
=> 
=> 
or 
., then 
, contradiction with (3)., so 
............(10)
for all 
.
for all 
, ............(11)
for all .
and use (11), then ![$[f(c)+1][2a]=2[f(ac)+a]$](http://latex.artofproblemsolving.com/a/c/6/ac6477d58375a279db595e36413c1495568f2163.png)
.
for all 
............(12)
for all 
, by taking 
., by (11) we have 
.![$2f(ac+a)=[f(c)+1]f(2a)=2[f(c)+1]f(a)$](http://latex.artofproblemsolving.com/7/2/3/7235d32582e516aef10f6035f5be8375ddcdd9a6.png)
=> 
.
, take 
, then 
.
for all => f is strictly increasing.........(13)
and 
, we can take some 
with 
.
while 
, false by (13).
, this is handled similarly.
for all 
, good.
for all 
. 
in that interval, we can find some 
with 
, so 
for these .
we get therefore if and then , Proving that is surjective. be a number s.t , then , so we get , so for . be any number, and sufficiently big numbers so , then: ?
satisfies:
by forgetting first few terms, plug in 
in the original equation we can get:![\[f(xf(a_n)+(1-x))<(a_n+1)f(1)<2f(1)\]](http://latex.artofproblemsolving.com/a/d/c/adc48660f204c82e5b56b8e5e1cb1bbefe959b5c.png)
Which tells us 
bounded by 
, take 
goes to infinite and get 
is bounded by , but if we take 
and very large in the original equation, RHS is smaller than 
, but LHS can be arbitrary large, hence we get a contradiction!
don't exist, hence 
exist.
, then we can find a sequence 
such that:
and take 
, we'll get![\[(z+1)k=\lim_{n\rightarrow\infty}(z+1)f(x_n)=\lim_{n\rightarrow\infty}2f(x_nf(z)+x_n)\leq 2k\]](http://latex.artofproblemsolving.com/6/5/9/6590c598df913e4d34dd40599fed7d2953c3dd5e.png)
So 
, and hence 
.
very small and 
very small and get a contradiction! is an increasing function.
and let 
we have
as 
and 
so 
and then 
and get 
such thatfor all positive real numbers . to find that![\[
(z+1)f(2x)=2f\big(xf(z)+x\big).\qquad (1).
\]](http://latex.artofproblemsolving.com/1/d/a/1da3fc6d783daa7fe602532d1f74894aede248d6.png)
Clearly, is injective, and running over 
in 
, we see that attains every value 
, 
, from which it can easily be deduced that 
, so is surjective. If 
, then gives 
; since is bijective, follows.
in yields 
. Plugging 
into this and rearranging gives 
whenever 
. Hence,![\[
\boxed{f(x)>\frac34\text{ if }x>\frac34.}
\]](http://latex.artofproblemsolving.com/b/7/8/b78d39241193c03929f35289d2ecad5e2c2e7061.png)
shows that 
is constant whenever 
. Due to surjectivity, is actually constant for all 
(and thus for all 
, too). Therefore, using , we get 
.
must be linear by Cauchy's equation, and so 
for some 
. Finally, from for 
, we get the identity 
. The finishing touches: injectivity and 
if imply 
; then taking the limit of both sides as 
shows 
. Therefore, for all , done. 
. Then throw in 
and 
. The RHS remains invariant, and the LHS changes, so this is a contradiction. Thus is injective.
to obtain 
. If 
, then is not injective, so .
and vary , we see that if 
, then we can attain any value in 
by varying , so is surjective.
. Then the original equation rearranges to 
. In particular, for 
we get Jensen's FE. Take 
with domain and range 
. Then we know 
. Observe that if 
for some 
, we would die because 
by additive, which contradicts the fact that the range is 
. Hence, 
for 
. Now if we take any 
, we get , but 
by injectivity so 
. This implies 
is increasing. Since is increasing and additive, it follows that it is linear by using the density of the rationals in the reals. is also linear. I'm too lazy to find the answer.
, which clearly is a solution to the equation. is injective. Indeed, setting 
and 
and supposing that 
gives that 
.
gives 
, so by injectivity . gives 
, so for any 
in the range of , 
is also in the range of for any . Surjectivity easily follows. and 
are distinct positive reals. Setting 
and 
with 
gives that 
. Taking with 
for a sufficiently small 
gives that 
for any 
with 
. and distinct that 
and 
. Adding gives 
, so 
for all 
. Thus, for any , so for any rationals 
and 
with 
and 
we have that 
. In other words, given two distinct numbers 
, 
will lie on the line through 
and 
provided that 
is a rationally weighted average of and . If such a line has a positive root, there is some interval in which it is negative, so we can rationally weight and into that interval resulting as 
is dense in 
resulting in a negative output of the function, a contradiction. Hence, all such lines must have nonpositive roots. In particular, this means that is increasing and that 
and 
.
and 
. Then, 
as 
, so 
. This means that 
for 
. Setting in gives that 
, so 
for all 
, implying that for all 
.
. This means that is surjective. gives only reals in 
. 
denote 
for all 
. is surjective on .
, 
and 
gives us
Now, for any 
, since is surjective, there exists 
that 
.
and 
, both are positive real numbers. There also exists 
that 
.
and 
.
gives us 
.
, there exists 
that 
for all 
.
, 
.
for all .
gives us 
for all 
., we get 
for all .
and 
for all . becomes
for all .
and 
for all 
.
for all 
.
for all 
.
, and similarly, 
for all .
that 
for all .
for all . In other words, 
for all 
.
be a function defined by 
for all .
is both multiplicative and additive, so must be identity function. So, 
for all . for all , which, hence, is the only solution.
denote the given assertion. is injective. is surjective.
is multiplicative. is involutive, i.e. 
is additive.
for all 
It is trivial to verify that this solution indeed satisfies the equation.
can take on arbitrarily small values.
By plugging in 
we know that:
Therefore, for any 
by letting 
, we can obtain another value 
such that 
From this, it is clear that we can get to tend arbitrarily close to 
, as desired. 
when 
. Consider fixing some for which is arbitrarily small. Then, observe that when we fix 
, we have that 
is fixed whenever 
and 
. Since is tiny, we can make such that 
and 
. Therefore, we are done.
now holds. Therefore, this furthermore implies that is nondecreasing by Jensen's Functional Equation. Since is 
it's easy to see that is linear. To see this, observe that for any two values and , then line through 
and 
has a dense subset which are points on 
Therefore, for any point 
which isn't on the line, we can clearly use the Jensen condition to get a negative value of (take a point 
on the line for which is arbitrarily close to on whichever side, depending on whether is over/under the line).Therefore, we obtain that 
. Plugging this into the equation, it's easy to see that only satisfies the desired conditions.
and 
. Here is the assertion that 
.
. Then 
gives 
. Now put 
to get a contradiction( basically getting that two unequal numbers are equal). Hence, .
.
with Claim 1 gives the result.
is multiplicative and surjective. can attain any value more than 
. Now, put 
to get 
or 
. As can obtain any value more than , we can see that if one of them is higher than 
. Now consider this:
. Then, obviously 
as 
is higher than . Now, 
which from previous result is same as even in the case when both are less than . Hence, for all reals. Now use this fact combined with that if 
, then there exists real with 
to prove that is surjective. with Claim 3 gives that 
. Now use being surjective to prove that for any given pair of positive reals 
one can find positive reals with 
which gives 
. is always a positive real), proving that should be trivial but I can't put it all together so this was all the progress I made.
is monotonically increasing. Suppose this is not true. One can find 
such that 
. One can see that 
. The LHS is always positive but the RHS eventually becomes negative, contradiction. Hence is monotonically increasing. From Jensen's FE and multiplicity one can obtain that 
where 
. Using that monotonically increasing, one can bound the value of for any real. One can show that 
is smaller than any given real for any hence for all .
such thatfor all positive real numbers .
for all positive reals .
denote the statement of the problem. and is injective.
for some positive reals and , then comparing 
and 
yields that 
or 
proving that is injective. Now see that 
implies that . is bijective, additive and multiplicative which essentially implies our desired result
, is bijective as we've already proved that it is injective. Now, see that if 
for some real 
, then 
and we see that 
holds true (using basic assertions) which means that in fact 
for which isn't difficult to see that it is true for 
too and so is multiplicative.
and 
, we yield that 
. Now, 
and now see that 
. If 
, then see that 
using basic assertions and if 
, it would contradict bijectivity and 
so 
and which immediately proves 
. Not hard to see that 
using all above assertions which means that is additive.
, 
for all positive reals which clearly satisfies the conditions of the given problem and we're done.
which obviously works. We now show that it is the only one. Label the equation
are positive reals then there exists positive reals such that
if and only if 
.
then 
if and only if is unbounded and injective. is bounded, then fixing and let tends to infinity, then 
is unbounded while 
is bounded, contradiction. Now, suppose 
, substituting 
and 
into we have
hence . and 
then
Proof. such that 
, by the lemma we can select 
such that 
, 
, 
, 
, then by we have
as desired. 
and so 
In particular, substituting in 
we have by injectivity. Substituting in and using ,
Hence substituting 
in 
we have 
, hence 
, solving
We have 
or 
, hence by injectivity, from we have 
, so for all positive integer , moreover from we have
Hence by Cauchy equation we can conclude that .
the assertion of the given F.E. is bijective
such that , thus by 
![$$f \left(\frac{2}{f(x)+1} \right)=\frac{2}{x+1} \implies f \; \text{surjective on} \; [0,2]$$](http://latex.artofproblemsolving.com/3/2/7/327c2a23708f6e0d16ebfa2f3f3317ea5f026ad9.png)
![$$f \left(\frac{f(x)+1}{2} \right)=\frac{x+1}{2} \implies f \; \text{surjective on} \; \left[\frac{1}{2}, \infty \right]$$](http://latex.artofproblemsolving.com/a/0/3/a03d1af60885c407144e13f809a7648dfcce8579.png)
Thus is bijective as desired. is multiplicative.
, by 
Now note that 
thus by 
Meaning that is multiplicative for all positive reals. is continuos
but we have:
Thus is continuos as desired.
and now let 
, we have:
The last step bc 
is also a solution to that but by Claim 1 we would have 
so 
which makes nonsense. we have that 
thus 
. Meaning that the unique sol is and we are done 
the assertion of the given F.E. is bijective such that , thus by ![$$f \left(\frac{2}{f(x)+1} \right)=\frac{2}{f(x)+1} \implies f \; \text{surjective on} \; [0,2]$$](http://latex.artofproblemsolving.com/f/6/4/f6478704d1a943aac5b4d6e174768804323e7ebb.png)
Thus is bijective as desired. is multiplicative., by Now note that thus by Meaning that is multiplicative for all positive reals. is continuos but we have:Thus is continuos as desired. and now let , we have:The last step bc is also a solution to that but by Claim 1 we would have so which makes nonsense. we have that thus . Meaning that the unique sol is and we are done 
to get that 
implies 
for all 
so we get surjectivity. And instead of continuity, we prove monotonicity, because something does not seem rigorous in the continuity part. How we prove monotonicity is by doing 
which gives us ![\[f(\frac{x(f(y)+1)}{2})=\frac{(y+1)f(x)}{2}\]](http://latex.artofproblemsolving.com/2/0/2/2023a4c87cb74d0faccb0254db50c59ca19009bb.png)
and we compare the domains on both sides. We get that ![\[f\left((\frac{x}{2},\infty\right)=(f(\frac{x}{2},\infty)\]](http://latex.artofproblemsolving.com/1/8/4/184ead4f47402991735bb2a04f17f38c2f00db55.png)
which proves our desired claim. From this we easily get 
because 
is multiplicative...
the assertion of the given F.E. is bijective such that , thus by Thus is bijective as desired. is multiplicative., by Now note that thus by Meaning that is multiplicative for all positive reals. is continuos but we have:Thus is continuos as desired. and now let , we have:The last step bc is also a solution to that but by Claim 1 we would have so which makes nonsense. we have that thus . Meaning that the unique sol is and we are done 
is continous not .
is the only solution to the FE which clearly works. Now we show this is the only solution. denote the given FE.
and gives us 
and we get 
for all real number . So actually is surjective function. with 
and with gives us 
. So has multiplicative property.
we get that 
. we can fix and choose such that 
and 
.
has additive property too. Since is additive and multiplicative, is the only solution.
So it is trivial that we may find the numbers 
Now 
and 
yield the claim. satisfies Jensen's Functional Equation, which we can solve (as we are in ) to get 
Checking gives 
We see that the identity satisfies.
is injective.
if 
then 
implies that 
and 
so 
. Assume 
then 
is bijective for all 
. 
is bijective for all just let be such that 
is near so our bound is being divided by each time and goes from 
to .
there exists 
such that 
. 
. 
so 
Note that 
so 
. from claim 
we have 
so 
and Note that 
so 
.
. Let 
. 
. 
so 
. Note that 
so 
. put 
and we have 
so 
and also 
so 
or 
and since then 
can't be 
so is . Also now we know that 
. 
is both additive and multiplicative so
is clearly injective
if we pin and slide along we get that 
and 
is surjective 
is bijective
and since is bijective 
such that 
bijective 
such that 
, 
is multiplicative
is an involution 
gives or 
but is injective 
is additive and multiplicative 
(since 
is impossible)
such thatfor all positive real numbers . be the assertion of the given functional equation.
, implying is surjective on 
since 
.
by the surjectivity. Substitute by 
into the above FE, we get 
Substitute by 
into , we get 
Subtracting from the two sides of and the above FE, we get 
so 
Comparing and with 
, we get 
Let and 
, then the above FE becomes 
so 
but is easily injective, thus 
be the assertion above.
is bijective.
Notice 
and 
for 
, easily gives us a contradiction, giving is injective. is surjective, notice gives us![\[(z+1)f(2x) = 2f(xf(z)+x)\]](http://latex.artofproblemsolving.com/6/5/a/65a83859ccf4ebfca8bf73b047ebb45a33eb02fd.png)
If 
, we get that 
has dimensions and can vary over all 
, giving a contradiction by setting arbitrarily large. Hence this gives us is surjective as desired. such that 
, taking 
gives us that or . is multiplicative. Now taking 
, we get![\[f(\frac{f(z)+1}{2}) = \frac{z+1}{2}\]](http://latex.artofproblemsolving.com/9/4/1/94153ca666d48dc73d54b16808ada4a055a622ca.png)
Now considering 
![\[f(\frac{xf(z)+x}{2}) = \frac{z+1}{2}f(x) = f(\frac{f(z)+1}{2})f(x)\]](http://latex.artofproblemsolving.com/c/a/5/ca5bdadb55e661fbad5cb5948a552bf15ad20427.png)
Pairing with the fact that is bijective, this gives us that is multiplicative where 
and 
. But notice if we consider 
and 
, we can take a 
to have![\[f(abc) = f(ab)f(c) = f(a)f(b)f(c) = f(ac)f(b)\]](http://latex.artofproblemsolving.com/e/c/2/ec284fbe3c6eec727e91af2696bba159d4addfb1.png)
Hence we get is multiplicative over all pairs of numbers. 
. Notice taking again and 
, we get that![\[f(\frac{z+3}{4}) = \frac{f(z)+3}{4}\]](http://latex.artofproblemsolving.com/d/7/8/d783b3a57560b67696bfac7883ed7e469ff2be6a.png)
with 
again, to have![\[4f(z+1) = f(4z+1) + 3\]](http://latex.artofproblemsolving.com/e/5/2/e52ebb27481be4ff14ef22211c9d94aed64b1225.png)
for all 
. Now taking , we get 
. This implies 
, which immediately gives us that 
as is bijective. Now taking 
gives . 
for all 
as is multiplicative. Also since 
, we also have 
for all . Taking 
we get![\[(2^i+1)f(x+y) = f(2^ix+y) + f(2^iy+x)\]](http://latex.artofproblemsolving.com/2/6/5/265123d506fbb743fe362cd395c97a377370e2de.png)
where 
. If we choose the pair![\[(x, y) = \left(\frac{2^ia-b}{2^{2i}-1}, \frac{2^ib-a}{2^{2i}-1}\right)\]](http://latex.artofproblemsolving.com/c/0/4/c0439267aa7b76bf579c8683616ba3e394277823.png)
where 
, we have 
. But note since this applies for all , we have is additive over all pairs of values. This clearly gives .
is bijective proof from Solution . Now notice first taking , we have that Imf
. Since is bijective, we have that then must be bounded over the interval 
. Hence we know for any sequence 
such that 
, we have that 
does indeed exist. Let this value be called 
.
such that 
is equal to the limsup subsequence, we get that![\[(z+1)M = f(xf(z)+y) + f(yf(z)+x)\]](http://latex.artofproblemsolving.com/f/a/d/fad05a477833a6ef44a6fae0c333faa003d6f70a.png)
The RHS is bounded, but taking to be an arbitrarily large value gives us a contradiction. Hence 
. Thus we have 
as 
., if we take 
, we can see that actually must be continuous on all values as left-hand side changes value only by linear 
and right-hand side has a change in input of 
. in original gives us Cauchy's Functional equation, concluding similarly to Solution that 
.
the assertion of the given functional equation. is bijective. then by 
we get that trivialy, now by 
and injectivity we get that , so now by 
And from we get that
And just to overkill this claim, by 
we get that
Hence is surjective and injective so its bijective as desired. is multiplicative.
then using Claim 1 we have 
takes all positive reals from , so now by 
we get that
Note it holds that 
due to how we defined and by 
we get that
Claim 3: 
for some real .
and Claim 3 we get that
Now this gives us 
so if 
then 
, so if u let 
then is additive and also if then 
so is bounded hence its linear so we get that 
or 
for each but if for some it held that 
then it holds that 
so either or 
, the last one also implying so holds for every positive real , thus we are done 
.
only, which clearly works. Let denote the assertion..
and suppose 
. Pick some with 
and let 
. Then 
, which means we can't have both 
and 
be at least : contradiction. satisfies Jensen's. and 
we obtain
Now for any positive reals , pick 
to be sufficiently small such that 
. Then by putting 
, we find that 
. is linear, and it is straightforward to check that only works. 
and varying we find that is surjective.
and varying we find that is surjective.
, then 
. includes all intervals 
, so is surjective. be any value such that . Then 
using 
, so 
.
. Then it follows, choose so that 
. Then 
.
it follows that there exist 
so that 
and plugging 
or 
:
.
, it follows that a shifted version of satisfies Cauchy's functional equation on 
and maps positive reals to reals bounded below by 
. In particular, it follows that is never negative, so is monotonic increasing, so is linear (the proof proceeds identically to the proof that Cauchy + bounded implies linear).
for 
. It follows from plugging large 
that 
for . Now reconsider 
for any . Sending to infinity, it follows that 
, so we have for all .
as the assertion of the equation in the problem. is additive over real positive, then 
for some constant 
. From claim 4, we get . Hence, 
![\[(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)\]](http://latex.artofproblemsolving.com/0/8/7/0878185d0795f50362613b440557762f483f62ed.png)
Only function holds the conditions is . Let be the assertion. is injective.
with 
gives![\[ (z+1)f(f(x)+1)=f(f(x)f(z)+1)+f(f(x)+f(z))=(x+1)f(f(z)+1)\]](http://latex.artofproblemsolving.com/c/8/a/c8a8f01b0ab45ad35792bb8c10d7eb888bcc9e67.png)
Choose to verify 
. Let 
. We observe that 
which proves the injectivity..
gives 
. Thus, 
. Since is injective, we conclude that .
. yields 
..![\[P(x,x,f(2z+1)+2): \ \ (f(2z+1)+3)f(2x)=4f(x(z+2))\]](http://latex.artofproblemsolving.com/c/f/1/cf1b7bea772b8dbc2d672eecad3e70b5cea65580.png)
Take 
to get 
. Note that this means 
. Now by using ,![\[f(f(\frac{c+1}{2-c})+1)=c(\frac{c+1}{2-c}+1)=\frac{3c}{2-c}=f(3)\implies f(\frac{c+1}{2-c})=2\]](http://latex.artofproblemsolving.com/b/3/1/b31e55ebdfba447677079c4633cac24778cc7890.png)
![\[P(\frac{c+1}{2(2-c)},\frac{c+1}{2(2-c)},z): \ \ f(\frac{f(2z+1)+1}{2})=z+1=f((f(z)+1)(\frac{c+1}{2(2-c)}))\]](http://latex.artofproblemsolving.com/7/7/a/77a0ca9b4823e780f97edbb50f5d6aa54a00944d.png)
Hence 
. Plugging gives 
.![\[P(\frac{1}{2},\frac{1}{2},\frac{f(5)+1}{2}): \ \ 2c=f(2)=\frac{f(5)+3}{4}\]](http://latex.artofproblemsolving.com/4/8/a/48a22cf16436eb91c53f16da4f7dc2edd0bcfbdc.png)
![\[\frac{(2c+1)(c+1)}{2-c}-1=f(5)=8c-3\implies 2c^2+3c+1-2+c=16c-6-8c^2+3c\iff (c-1)^2=0\iff c=1\]](http://latex.artofproblemsolving.com/4/f/e/4fee10e8a07fbba78cd6c2f556cbfb056da40eff.png)
Which completes the proof. is additive.
.![\[P(x,y,f(z)+1): \ \ (f(z)+2)f(x+y)=f(xz+x+y)+f(yz+x+y)\]](http://latex.artofproblemsolving.com/7/6/d/76d06948dd0e89b2d4f8b69f57eec5d153a79821.png)
Plugging 
yields 
.
for all positive integers. 
implies 
hence 
.![\[P(x,x,2): \ \ 3f(2x)=2f(3x)=6f(x)\implies f(2x)=2f(x)\]](http://latex.artofproblemsolving.com/b/1/9/b198f1976006dcb27744c18606fd4d9ce13dc3d7.png)
gives 
thus, 
. So![\[f(xz+yz+2x+2y)=f(z+2)f(x+y)=f(xz+x+y)+f(yz+x+y)\]](http://latex.artofproblemsolving.com/1/d/b/1db8a6e09d33ddc3da813639048619e01a398878.png)
We claim that for all 
, we can find such that 
and 
. For 
we proved that 
. We can assume that 
.![\[b-x-yz=y=a-x-xz\implies z=\frac{a-b}{x-y}\]](http://latex.artofproblemsolving.com/4/b/7/4b7691a71e26e202084133d9b9db6790670d0d14.png)
. We choose 
.![\[x=\frac{b-\sqrt{4y^2-4ya+b^2}}{2}>\frac{b-\sqrt{(b-2y)^2}}{2}=y\]](http://latex.artofproblemsolving.com/7/3/7/7373f45ec6de1a3893e3ecf9ba06db5ae7ba5c51.png)
So . Since there exists for all we observe that 
hence is additive.![\[(z+1)f(x)+(z+1)f(y)=f(xf(z))+f(yf(z))+f(x)+f(y)\iff zf(x)+zf(y)=f(xf(z))+f(yf(z))\]](http://latex.artofproblemsolving.com/f/7/e/f7e21c0efc52784b74af506c02033d776884b485.png)
Plug 
to get (by using 
![\[(f(z)+1)f(x+2)+(f(z)+1)f(y)=f(xz+x+2z+2)+f(yz+y)\]](http://latex.artofproblemsolving.com/a/5/3/a53c14dfe79cfda9a18565ae523c1a85191df972.png)
![\[2f(z)+f(x)+f(y)f(z)+f(y)=f(x)+2f(z)+f(yz)+f(y)\iff f(y)f(z)=f(yz)\]](http://latex.artofproblemsolving.com/4/c/4/4c4924a4c31d47c9f5cb022be24007efd3d111e4.png)
Thus, is multiplicative. Since is both additive and multiplicative, we get that as desired.
and 
.
![\[(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x).\]](http://latex.artofproblemsolving.com/1/0/f/10f982d24bc662f523a579e1f14435a59c3e7b4f.png)
.
,
,
.
,![\[(z+1)f(2x)=2f(xf(z)+x)\implies f(2x)\cdot\frac{z+1}{2}=f(xf(z)+x).\]](http://latex.artofproblemsolving.com/b/a/a/baa41aaa536764525e2f6fb2bb6f90814494b847.png)
Thus, if 
.
,
,
,
,
.
.
satisfying
then we would be done since![\[f(a)+f(b)=(z+1)f(x+y)=(z+1)f\left(\frac{a+b}{f(z)+1}\right)=(z+1)f\left(\frac{a'+b'}{f(z)+1}\right)=(z+1)f(x'+y')=f(a')+f(b').\]](http://latex.artofproblemsolving.com/9/0/c/90c267eb42c0457a253f47a5269011ea1fd3cb2b.png)
Since 
.
)![\[x=\frac{ka-b}{k^2-1},y=\frac{kb-a}{k^2-1},x'=\frac{ka'-b'}{k^2-1},y'=\frac{kb'-a'}{k^2-1}.\]](http://latex.artofproblemsolving.com/d/e/d/ded94cdca70157b4f6b0c5a6bab7ebcd617d96f8.png)
Letting 
are all positive, as desired. Hence, 
.![\[f(a)+f(b)=2f\left(\frac{a+b}{2}\right).\]](http://latex.artofproblemsolving.com/c/c/5/cc5824ab557769c181554755af57ed7a37b01477.png)
If 
.
.
Adding these equations and subtracting 
from both sides gives 
as desired.
by 
for all 
.
,
for all 
.
,
;
,
for all 
and 
by an easy induction. In addition, 
for all 
was such that 
.
,
for all 
,
,
.
for all 
.![\[g(x)+g(1)=2g\left(\frac{x+1}{2}\right)\implies g(x)+c=(x+1)c\implies g(x)=cx.\]](http://latex.artofproblemsolving.com/5/0/8/5088edbec4b18ff40d3ace9b90bdff0ea3ce66e7.png)
Hence, 
for all 
.
,![\[2f\left(\frac{1}{2}f(z)+\frac{1}{2}\right)=2f\left(\frac{1}{2}(c(z-1)+1)+\frac{1}{2}\right)=2f\left(\frac{1}{2}c(z-1)+1\right)\]](http://latex.artofproblemsolving.com/a/4/3/a4385eaf5f164fae27cf9bb5ec22b5006d474238.png)
![\[=2\left[c\left(\frac{1}{2}c(z-1)\right)+1\right]=c^2(z-1)+2=c^2z-c^2+2.\]](http://latex.artofproblemsolving.com/3/1/c/31ca8b2bfc46874ce70121e61256b62f8eca91af.png)
Hence, 
,
.
.
corresponds to 
,
and 
,
is positive then we conclude that 
since 
,
(let the assertion of this one is 
)
,
there exists 
.
then fixed that 
there exists 
(take this limit we can say this is still true forall 
)
,
holds 
.
holds if 
.
,
.
.
forall 
,
large enough satisfy 
then we get 
.
,
.
.
.
.
,
such that 
and 
.
.
.
.
.
and 
.
.
.
in 
.
.
.
.
for all 
.
and 
implies 
.
.
then this implies that 
for 
.
with 
.
implies 
and hence 
with 
.
implies 
and hence 
for all 
let there exists a unique 
with 
.
implies 
and hence 
holds for all 
and hence holds for all 
i.e. 
.
implies 
and hence by multiplicativity![\[\frac{(z+1)f(2)}{2}=f(f(z)+1)=f(f(z)) \cdot f(1+\frac{1}{f(z)})=f(f(z)) \cdot f(f(\frac{1}{z})+1)=f(f(z)) \cdot \frac{(z+1)f(2)}{2z}\]](http://latex.artofproblemsolving.com/c/7/f/c7fdcac2044284087e5c21ddb2c72f9fc102c459.png)
and hence 
and hence 
where 
.
and hence 
.
then ![\[f(x+y)=f(x) \cdot f(1+\frac{y}{x})=f(x) \cdot (1+f(\frac{y}{x})=f(x)+f(x) \cdot f(\frac{y}{x})=f(x)+f(y)\]](http://latex.artofproblemsolving.com/3/8/c/38c0ee2e569024d73a098e447ccfaec2c2971886.png)
i.e. 
denote the statement of the functional equation. For 
denote 
.
are such that 
and 
we get 
.
gives 
.
.
and 
.
If we let ![$]\tfrac{1}{2},+\infty[$](http://latex.artofproblemsolving.com/a/a/5/aa5ed6681158e496cb710c72ede7a2f542c1a55e.png)
and ![$]0,2[$](http://latex.artofproblemsolving.com/c/d/8/cd8f01afee9842395952229c046f5b7d9bcb4a46.png)
,
.
,
If we fix 
runs through ![$]\tfrac{f(x)}{2},+\infty[$](http://latex.artofproblemsolving.com/0/2/8/02830ce5dbdeece565aafe679bd1a1ad6c28e9a7.png)
while 
runs through ![$]\tfrac{x}{2},+\infty[$](http://latex.artofproblemsolving.com/4/7/e/47e4d16a5aec752c400e273b68eb642f415f09fc.png)
.![\begin{align*}
\boxed{f\left(\Big]\frac{x}{2},+\infty\Big[\right)=\Big]\frac{f(x)}{2},+\infty\Big[}.
\end{align*}](http://latex.artofproblemsolving.com/2/0/0/200dc2499f4b6fc9f59c46f56972ee5aaa7e68a0.png)
be such that 
.
implies ![$]\tfrac{x}{2},+\infty[\;\subset\;]\tfrac{y}{2},+\infty[$](http://latex.artofproblemsolving.com/4/1/d/41d819c6ed30892db736c7fbcc681fc0bccb6d1a.png)
.![$f(]\tfrac{x}{2},+\infty[)\;\subset\;f(]\tfrac{y}{2},+\infty[)$](http://latex.artofproblemsolving.com/e/d/6/ed6d9203bf84d0da516f30538ec8fb8258a9762f.png)
,![$]\tfrac{f(x)}{2},+\infty[\;\subset\;]\tfrac{f(y)}{2},+\infty[$](http://latex.artofproblemsolving.com/6/0/6/6063db29a680670970963ec98f4edf323d80ccb2.png)
.
.
.
for 
(and 
for 
)![$f(]1,+\infty[)=]1,+\infty[$](http://latex.artofproblemsolving.com/e/b/2/eb23b2774352ab7f869b4d1c8ff8157cd686ab43.png)
.![$f(]1,+\infty[)=]\tfrac{f(2)}{2},+\infty[$](http://latex.artofproblemsolving.com/2/9/1/2915921bf8e2f2c83bacf6e8c85f8c484aaf7b20.png)
,
,
.
and it follows easily by induction that 
.
define 
and 
.
are fixed points of 
which is just 
are fixed points of 
yields
and since 
are also fixed points of 
and 
.
and plugging in ![$z\in\;]1,+\infty[$](http://latex.artofproblemsolving.com/5/0/2/50230c32bf6e4e5a569efdb58959171df119aa4f.png)
we obtain also 
,
which is clearly a solution.
is injective.
By LEMMA-1, 
is surjective 
.
is surjective 
.
in 
,
and 
,
,
,
,
,
,
,
,
,
in 
,
,
![$y = \frac{z+3}{4} \Rightarrow y>\frac{3}{4} \Rightarrow y>\frac{1}{2} \Rightarrow f(f(y)) = y \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } [Using \text{ } (I)]$](http://latex.artofproblemsolving.com/c/b/d/cbd414f4860a767d2eb4c65d7e963502e238b25d.png)
![$\Rightarrow f\left(f\left(\frac{z+3}{4} \right) \right) = \frac{z+3}{4} \Rightarrow f\left(\frac{f(z)+3}{4} \right) = \frac{z+3}{4} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } [Using \text{ } (G)]$](http://latex.artofproblemsolving.com/2/b/1/2b1a5b8e41ade16ead7e282600f65c7d6101c6ac.png)
LEMMA-![$]$](http://latex.artofproblemsolving.com/0/7/1/07174a92e49be0bb0db1beef2bb6c245a8b212f9.png)
,
and 
,
,
or 
or 
,
,
,
,
for some 
is additive.
and 
actually we get 
implies 
gives 
.
Now, using equation 1 in 
gives us
Replacing 
with 
in equation (2), we get
Again replacing 
in equation (2), we get
,
.
but we have if 
then 
.
.
,
Substituting 
gives
By substituting 
.
,
.
.
and plugging the new equation back we get 
and multiplicavity, we get
By comparing it to 
must be equal to 
and the fact that 
.
.
,
.
.
,
and 
are real positive for every positive real numbers 
.
,
Because 
and by multiplicavity, we get 
for every positive real numbers 
.
and 
,