My solution: Let $P(x,y):f(x+f(x)+f(y))=2f(x)+y$ $P(d,y)$ implies f is injective and surjective on $[2f(d),+\infty)$ where $d$ is constant. $P(x,2f(y))$ implies that $Q(x,y):f(x+f(x)+f(2f(y))=2f(x)+2f(y)$. Now $Q(x,y)vsQ(y,x)$ gives $f(x+f(x)+f(2f(y)))=f(y+f(y)+f(2f(x)))$ thus $f(2f(x))=x+f(x)+c$(1) by the injectivity of $f$. Hence $P(x,y)$ rewrites as $f(f(2f(x))-c+f(y))=2f(x)+y$ and since $f(x)$ is surjective on $[2f(d),\infty)$ we have that $f(f(x)+f(y)-c)=x+y$ for all large x and $y\in\mathbb{R+}$. Setting $x\to f(x)+f(y)-c$ (x large) in (1) we obtain that $f(2(x+y))=f(x)+x+f(y)+y$ for all x large and $y\in \mathbb{R+}$. For all x,y,z large, consider $P(x,2(y+z))\Rightarrow Q(x,y,z): f(x+f(x)+f(y)+y+f(z)+z)=2f(x)+2(y+z)$. Now we compare $Q(x,y,z)$ with $Q(y,x,z)$ to get that $f(x)-x=f(y)-y$ for all large $x,y$. Thus $f(x)=x+c$ for all large x. Taking a large $x$ in (1) we conclude that c=0 implying that $f(x)=x$ for all large x. $P(x,y)$ where $x$ is large implies that $f\equiv id$ for all $x \in\mathbb{R+}$. We can easily check that this function satisfies the given FE.

Here is a solution using only substitutions. Let $P(x,y)$ denote the assertion that $$f(x+f(x)+f(y)) = 2f(x)+y.$$ $P(x,x)$ gives us: $$f(x+2f(x)) = 2f(x)+x.$$ Using the above, $P(x+2f(x), x)$ gives us: $$f(x+2f(x) + x + 2f(x)+f(x)) = 2(x+2f(x))+x$$ which is equivalent to: $$f(2x+5f(x)) = 3x+4f(x).$$ Similarly, from $P(x,x+2f(x))$ we get: $$f(x+f(x)+x+2f(x)) = 2f(x)+x+2f(x)$$$$ \iff f(2x+3f(x)) = x+4f(x).$$ Now using the above equation, $P(x,2x+3f(x))$ gives us: $$f(x+f(x)+x+4f(x)) = 2f(x)+2x+3f(x)$$$$\iff f(2x+5f(x))=2x+5f(x).$$ Finally, we have: $$2x+5f(x) = f(2x+5f(x)) = 3x+4f(x)$$which means that $f(x)=x$.

Solution 1. $$P(x,x) : f(2f(x)+x)=2f(x)+x$$2. $$P(x,2f(x)+x): f(3f(x)+2x)=4f(x)+x$$(from 1) 3. $$P(x,3f(x)+2x) : f(5f(x)+2x)=5f(x)+2x$$(from 2) 4. $$P(2f(x)+x,x): f(5f(x)+2x)=4f(x)+3x$$(from1) So finally $$4f(x)+3x=5f(x)+2x\Rightarrow f(x)=x$$for all positive reals

As above $f$ is injective and surjective on $(2f(1),+\infty)$. Also $t=1+2f(1)$ is a fixed point, and so $f(y+4t)=f(2t+(2t+y))=f(2t+2f(t)+y)=f(2t+f(t+f(t)+f(y)))=f(t+f(t)+f(2t+f(y))=2f(t)+2t+f(y)=f(y)+4t$ Also by induction we have $f(y+4nt)=f(y)+4nt$. Now, since $f$ is surjective for big enough values, take $N$ so that $4Nt-f(y)>2f(1)$, and take a $x$ such that $f(x)=4Nt-f(y)$. So we have $8Nt-f(y)=f(x)+4Nt=f(x+4Nt)=f(x+f(x)+f(y))=2f(x)+y=2(4Nt-f(y))+y\implies f(y)=y$ Which clearly always works

Soundricio wrote: My solution: I am sure that the proposer is dangerousliri. No, the proposer is Socrates!

Let $P(x,y)$ denote the given assertion. $P(x,x)\Rightarrow f(x+2f(x))=x+2f(x)$ $P(x,x+2f(x))\Rightarrow f(2x+3f(x))=x+4f(x)$ $P(x,2x+3f(x))\Rightarrow f(2x+5f(x))=2x+5f(x).$ $P(x+2f(x),x)\Rightarrow 2x+5f(x)=3x+4f(x)\Rightarrow\boxed{f(x)=x}$ which works.

Soundricio wrote: I am sure that the proposer is dangerousliri Why if a functional is to BMO, should be mine. Others propose good functionals too. So, I congrat socrates for this good problem.

Let $P(x,y)$ be $ f(x+f(x)+f(y))=2f(x)+y$ (1) $P(a+f(a)+f(b),y)$ gives: $f(a+f(a)+f(b)+f(a+f(a)+f(b))+f(y))=2f(a+f(a)+f(b))+y$ or $f(a+f(a)+f(b)+2f(a)+b+f(y))=4f(a)+2b+y$ Let this be $R(a,b,y)$ $P(x,c+f(c)+f(d))$ gives: $f(x+f(x)+f(c+f(c)+f(d)))=2f(x)+c+f(c)+f(d)$ or $f(x+f(x)+2f(c)+d)=2f(x)+c+f(c)+f(d)$ Let this be $Q(x,c,b)$ $Q(x,x,f(x)+x+f(y))$ gives: $f(2x+4f(x)+f(y))=3f(x)+x+f(f(x)+x+f(y))$ (1) $R(x,x,y)$ gives:$f(2x+4f(x)+f(y))=4f(x)+2x+y$. (2) (1) and (2) gives:$4f(x)+2x+y=3f(x)+x+f(f(x)+x+f(y))$or $4f(x)+2x+y=3f(x)+x+2f(x)+y$ or $f(x)=x$. Edit: Actually I have get: $P(x+2f(x),y)$ and $P(x,x+f(x)+f(f(x)+x+f(y)))$

First prove injectivity. Then, one can show that for any nonnegative integer $k$ and a positive real $x$, $$f((3k+2)f(x)+(k+1)x)=(3k+2)f(x)+(k+1)x$$and $$f((3k)f(x)+(k+1)x)=(3k+1)f(x)+kx$$by induction on $k$. Then, just plug $P((3k+2)f(x)+(k+1)x, (3k)f(x)+(k+1)x)$ and finish.

VicKmath7 wrote: Find all functions $f:R+$ $\rightarrow$ $R+$, such that $f(x+f(x)+f(y))=2f(x)+y$ for all positive reals $x,y$. Easy and nice. $P(1,y)\implies f(1+f(1)+f(y))=2f(1)+y$ $P(x,1+f(1)+f(y))\implies f((x+f(x)+2f(1))+y)=(2f(x)+1+f(1))+f(y)$ This Lemma$\implies \frac{2f(x)+1+f(1)}{x+f(x)+2f(1)}=const\to f-linear$ Then $f(x)=x$.

We claim that the only possible solution is $\fbox{f(x)=x}$ Let $P(x,y)$ be our assertion then, Claim 1:-$f$ is injective. $P(a,a) \rightarrow$ $$f(a+2f(a))=2f(a)+a$$$P(a,b) \rightarrow $ $$f(a+f(a)+f(b))=2f(a)+b$$let $f(a)=f(b)$, then $f(a+f(a)+f(a))=f(a+f(a)+f(b)) \Rightarrow 2f(a)+a=2f(a)+b \Rightarrow a=b$ Hence, $f$ is injective as desired. $P(x,x) \rightarrow $ $$f(x+2f(x))=2f(x)+x$$$P(x,x+2f(x)) \rightarrow $ $$f(2x+3f(x))=4f(x)+x$$$P(x,2x+3f(x)) \rightarrow $ $$f(2x+5f(x))=2x+5f(x)$$$P(x+2f(x),x) \rightarrow $ $$f(x+2f(x)+x+2f(x)+f(x))=f(2x+5f(x))=3x+4f(x)$$$\Rightarrow 2x+5f(x)=3x+4f(x) \Rightarrow \fbox{f(x)=x}$

Note that $f(1+2f(1))=1+2f(1)$. Hence, $f$ has a fixed point. Call it $t$. Consider an arbitrary positive real number $y$. Then $P(t, y)$ gives us $$f(f(y)+2t)=y+2t.$$Furthermore, $P(t, f(y)+2t)$ gives us $$f(y+4t)=f(y)+4t.$$Finally, $P(f(y)+2t, 1)$ gives us $$f(f(y)+y+4t+f(1))=2(y+2t)+1.$$However, $$f(f(y)+y+4t+f(1))=f(f(y)+y+f(1))+4t=2f(y)+1+4t.$$Hence, $2(y+2t)+1=2f(y)+1+4t$, which means $f(y)=y$. Since $y$ was arbitrary, the indentity function is the only candidate for the solution, and it's easy to check that it is a solution.

Even though the problem got praise from others, I strongly dislike the problem and generally functional equations of this flavor. The sad part about the problem is that it is not actually an $\mathbb{R}^{+}$ FE, nevertheless here is my solution. Let $P(x,y)$ denote the above assertion. Notice by $P(x,x)$ that $$f(x+2f(x))=x+2f(x) $$Then by $P(x,x+2f(x))$, we have that $$f(2x+3f(x)) = 4f(x)+x$$which by $P(x,2x+3f(x))$ and $P(x+2f(x),x)$ gives us that $$3x+4f(x) = 2f(x+2f(x))+x = f(2x+5f(x)) = 2f(x)+2x+3f(x) = 2x+5f(x)$$meaning that $f(x) = x$ for all $x \in \mathbb{R}^{+}$. $\blacksquare$

Much easy

Do I win the least substitution contest? $P(x,x+f(x)+f(y)+f(x+f(x)+f(y))+f(f(y))): f(f(y))=y$ $P(f(x),y): f(x)=x$. All verification and all the pain is left to the reader as an exercise.

you must first prove that f is greater than some number, then f is easily found for numbers greater than this number.

Does this work?

I am wondering since the last substitution doesn't seem to have been used.1

3:38 Let $P(x,y)$ denote the given assertion. $P(x,x): f(x+2f(x))=x+2f(x)$. $P(x+2f(x),x): f(2x+5f(x))=3x+4f(x)$. $P(x,x+2f(x)): f(2x+3f(x))=x+4f(x)$. $P(x,2x+3f(x)): f(2x+5f(x))=2x+5f(x)$. Thus, $2x+5f(x)=3x+4f(x)\implies \boxed{f(x)=x}$, which works.

Ans: $f(x)=x$ for all positive real $x$. Pf: Let $P(x,y)$ denote the given assertion, \[P(x,x) \implies f(x+2f(x))=x+2f(x)\]\[P(x+2f(x),x)\implies f(2x+5f(x))=2x+5f(x)\]\[P(x,2x+5f(x))\implies f(3x+6f(x))=2x+7f(x)\]\[P(x+f(x),x+f(x))\implies f(3x+6f(x))=3x+6f(x)\]so \[2x+7f(x)=3x+6f(x) \implies f(x)=x. \ \ \blacksquare\]

$P(x,x) : f(2f(x) + x) = 2f(x) + x$ $P(x,2f(x) + x) : f(3f(x) + 2x) = 4f(x) + x$ $P(x,3f(x) + 2x) : f(5f(x) + 2x) = 5f(x) + 2x$ $P(2f(x) + x,x) : f(5f(x) + 2x) = 4f(x) + 3x$ so $4f(x) + 3x = 5f(x) + 2x \implies f(x) = x$

Alternatively, by this lemma, $2f(x)-f(x)-x=f(x)-x$ is constant, so $f$ is linear, and only $f(x)=x$ works.

The solution is $f(x) \equiv x$, which clearly works. Let $P(x,y)$ be the given assertion. Claim 1: $f$ is injective. Proof: If $f(a) = f(b)$, compare $P(x,a)$ and $P(x,b)$ to obtain $a=b$. $\square$ Call a pair $(r,s)$ of real numbers good if $$ f(x+r) = f(x) + s ~~ \forall ~ x > 0,-r \qquad \qquad (1)$$As $f$ takes only positive values, so it isn't hard to see that $r,s$ must have the same sign. Also, as $f$ is injective, so either of $r,s$ being equal to $0$ implies the other to equal $0$ too, i.e. if $rs=0$ then we must have $(r,s) = (0,0)$. Claim 2: For any $x_1,x_2 \in \mathbb R^+$, for $$ r = 2f(x_1) - 2f(x_2) ~~~,~~~ s = (f(x_1) + x_1) - (f(x_2) + x_2) $$the pair $(r,s)$ is good. Proof: Pick any $y_1,y_2 \in \mathbb R^+$ satisfying $$ 2f(x_1) + y_1 = 2f(x_2) + y_2 $$Compare $P(x_1,y_1)$ and $P(x_2,y_2)$. Using injectivity of $f$ we obtain $$ x_1 + f(x_1) + f(y_1) = x_2 + f(x_2) + f(y_2) $$It isn't hard to see that our Claim follows. $\square$ Claim 3: If $(r,s)$ is good, then $(2s,r+s)$ is also good. Proof: Observe that $$ 2f(x+r) - 2f(x) = 2s ~~~,~~~ (f(x+r) + x+r) - (f(x) + x) = r+s $$So we are done by Claim 2. $\square$ Claim 4: $f$ has arbitrarily large fixed points. Proof: $P(x,x)$ gives $$f(2f(x) + x) = 2f(x) + x$$As $f(x) \in \mathbb R^+$, so as $x$ becomes large, $2f(x) + x$ also becomes large, as desired. $\square$ Claim 5: If $(r,s)$ is good, then we must have $|r| \ge |s|$. Proof: By definition, $(r,s)$ is good iff $(-r,-s)$ is good. We may assume $r,s$ are positive now (case $rs=0$ is direct). We want to show $r > s$. Pick a large $t$ for which $f(t) = t$. Pick largest $k \in \mathbb Z_{\ge 0}$ for which $kr < t$. Then, $$0 < f ( t - kr) = f(t) - ks = t - ks \le (k+1)r - ks$$As $t$ can become arbitrarily large, so $k$ can become arbitrarily large, which implies our Claim. $\square$ Claim 6: If $r,s$ is good, we must have $r=s$. Proof: We may assume $r,s > 0$. By Claim 3 we know $(2s,r+s)$ is also good. Using Claim 5 we obtain $$ r \ge s ~~,~~ 2s \ge r+s $$This forces $r=s$, as desired. $\square$ We are ready to finish. Using Claim 2 and Claim 6 we obtain that for $x_1,x_2 \in \mathbb R^+$, $$ 2f(x_1) - 2f(x_2) = (f(x_1) + x_1) - (f(x_2) + x_2) \implies f(x_1) - f(x_2) = x_1 - x_2 $$This gives that $$ f(x) \equiv x+c $$for some constant $c \in \mathbb R$. We must have $c=0$, say because: Just by Plugging in $f(x) \equiv x+c$ in $P(x,y)$. $f$ as a fixed point by Claim 4. This completes the proof. $\blacksquare$

Nice Problem ! Let $P(x,y) : f(x+f(x)+f(y))=2f(x)+y$ $P(x,x) : f(x+2f(x))=2f(x)+x$ ( ) By ( ) and $P(x,x+2f(x))$ we get $f(2x+3f(x))=4f(x)+x$ ( ) By ( ) and $P(x,2x+3f(x))$ we get $f(2x+5f(x))=5f(x)+2x$ ( ) $P(x+2f(x),x) : f(5f(x)+2x)=4f(x)+3x)$ . ( ) Now by ( ) and ( ) we get our function is $f(x)=x$

Look this function for P(x,x)

Let $P(x,y)$ be the given assertion. We now proceed with the basic substitution of $P(x,x)$, which gives: $$f(2f(x) + x) = 2f(x) + x$$$P(x, 2f(x) + x)$ gives: $$f(3f(x) + 2x) = 4f(x) + x$$$P(2f(x) + x, x)$ gives: $$f(5f(x) + 2x) = 4f(x) + 3x$$$P(x, 2x + 3f(x))$ gives: $$f(5f(x) + 2x) = 5f(x) + 2x$$This means that $5f(x) + 2x = 4f(x) + 3x \implies f(x) = x$ so we are done.

We claim that the only solution is $\boxed{f\equiv \text{id}}$, which clearly works. We now show that it is the only one; let $P(x,y)$ denote the assertion. Claim: $f$ has arbitrarily large fixed points. Proof: From $P(x,x)$ we have that $f(2f(x)+x) = 2f(x)+x$, which implies $f$ has a fixed point. Suppose $a$ is a fixed point of $f$. Then, $P(a,a)$ gives $f(3a)=3a$, which means $3a$ is also a fixed point of $f$. By induction, $3^na$ is a fixed point of $a$ for any positive integer $n$, which proves our claim. Claim: $f$ is surjective on some interval $(r,\infty)$, where $r\in \mathbb{R}^+$. Proof: Fix $x$. Then, note $f(x+f(x)+f(y)) = 2f(x)+y$ can take on any positive real value greater than $2f(x)$ simply by varying $y$; hence proved. Now, let $x$ be any positive real. From our two claims, we know there exists some positive real $y$ such that $x+f(x)+f(y)$ is a fixed point of $f$. Fix $y$ as such a value; then, we must have $x+f(x)+f(y) = 2f(x)+y$, or $f(x)-x = f(y)-y$, which implies $f(x) = x+c$, for some nonnegative real $c$. Plugging this into the given equation yields $2x+y+3c = 2x+y+2c$, so $c=0$, as desired. $\blacksquare$

Denote $P(x,y)$ as the assertion of the following F.E. If $f(a)=f(b)$ then by $P(x,a)-P(x,b)$ gives $f$ injective as there is one term $y$ on RHS. $P(x,x)$ gives $f(x+2f(x))=x+2f(x))$ $P(x+2f(x),x)$ gives $f(2x+5f(x))=3x+4f(x)$ $P(x,x+2f(x)$ gives $f(2x+3f(x))=x+4f(x)$ $P(x,2x+3f(x))$ gives $f(2x+5f(x))=2x+5f(x)$ Therefore joining both gives we get $f(x)=x$ for all positive reals $x$ as desired, thus we are done .

Write $(a,b) \rightarrow (c,d)$ to mean $f(af(x)+bx)=cf(x)+dx$ for all $x$. First $x=y$ gives $(2,1)\rightarrow (2,1)$, and so $2f(x)+x$ is a fixed point. Next note that if $a$ is a fixed point, then $x=y=a$ gives $3a$ is also a fixed point. So $(6,3) \rightarrow (6,3)$ too. But set $y=af(x)+bx$ into the original equation means $(a,b)\rightarrow (c,d) \Rightarrow (c+1,d+1)\rightarrow (a+2,b)$. Now, $(2,1)\rightarrow (2,1) \Rightarrow (3,2)\rightarrow (4,1) \Rightarrow (5,2) \rightarrow (5,2) \Rightarrow (6,3) \rightarrow (7,2)$ So $f(6f(x)+3x)=7f(x)+2x=6f(x)+3x \Rightarrow f(x)=x$.