Here is my solution:

Setting $x\mapsto \frac1x$ in the given equation gives $$f\left(f(x)+\frac1y\right)=x+\frac1{f(y)}.$$Call this statement $P(x,y)$. This immediately gives $f$ is injective. Now comparing $P(x,\tfrac1{f(y)})$ and $P(y,\tfrac1{f(x)})$ gives $$\frac{1}{f\left(\frac1{f(x)}\right)}-\frac{1}{f\left(\frac1{f(y)}\right)}=x-y\implies \frac{1}{f\left(\frac1{f(x)}\right)}=x+k\implies f\left(\frac1{f(x)}\right)=\frac1{x+k}.$$Here $k$ is some constant. Also, comparing $P(\tfrac1{f(x)},y)$ and $P(\tfrac1{f(y)},x)$ and using injectivity, we have $$f\left(f\left(\frac1{f(x)}\right)+\frac1y\right)=f\left(f\left(\frac1{f(y)}\right)+\frac1x\right)\implies f\left(\frac1{f(x)}\right)+\frac1y=f\left(\frac1{f(y)}\right)+\frac1x\implies f\left(\frac1{f(x)}\right)=\frac1x+k'.$$Here $k'$ is another constant. Now this gives $\tfrac1{x+k}=\tfrac1x+k'$ holds for all $x\in\mathbb R^+$, which forces $k=k'=0$. So in fact $f\left(\frac{1}{f(x)}\right)=\frac1x.$ Now $P(\tfrac{1}{f(x)})$ gives $$f\left(\frac1x+\frac1y\right)=\frac1{f(x)}+\frac1{f(y)}.$$Call this new statement $Q(x,y)$. Now $Q(x,x)$ gives $f(\tfrac2x)=\tfrac2{f(x)}\;(\star).$ Comparing $Q(\tfrac{xy}{x+y},1)$ and $Q(x,\tfrac{y}{y+1})$ and mutilplying y $2$ gives $$\frac2{f(1)}+\frac{2}{f\left(\frac{xy}{x+y}\right)}=\frac{2}{f\left(\frac y{y+1}\right)}+\frac2{f(x)}.$$Using $(\star)$ in each of these terms, we have $f(2)+f\left(\tfrac2x+\tfrac2y\right)=f\left(2+\tfrac{2}{y}\right)+f\left(\tfrac2x\right)$, and replacing $x\mapsto 2x,y\mapsto 2y$, we get $$f(2)+f\left(\frac1x+\frac1y\right)=f\left(2+\frac{1}{y}\right)+f\left(\frac1x\right).$$Now we use the statements $Q(x,y),Q(\tfrac12,y)$ to simplify that into $$f(2)+\frac1{f(x)}+\frac{1}{f(y)}=\frac1{f(\tfrac12)}+\frac1{f(y)}+f\left(\frac1x\right)\implies f\left(\frac1x\right)-\frac1{f(x)}=f(2)-\frac1{f(\tfrac12)}.$$Setting $x=2$, there, we get $f(\tfrac12)+\tfrac1{f(1/2)}=f(2)+\frac1{f(2)}$, and since $f(2)\ne f(1/2)$ because of injectivity, we get $f(2)=\frac1{f(1/2)}$, which in turn implies $f\left(\frac1x\right)=\frac1{f(x)}.$ Now $Q(x,y)$ can be written as $f(\tfrac1x+\tfrac1y)=f(\tfrac1x)+f(\tfrac1y)$, which becomes Cauchy's equation after setting $x\mapsto 1/x,y\mapsto 1/y$. Since the codomain of $f$ is bounded from below, we must have $f(x)=cx$, and only $f(x)=x$ fits. Edit: Sniped darn

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We can rewrite it as: $$\frac{1}{x}+\frac{1}{f(y)}=f(\frac{1}{y}+f(\frac{1}{x})).$$It is clear that we can replace $x$ by $\frac{1}{x}$ and get: $$x+\frac{1}{y}=f(\frac{1}{y}+f(x)).$$Suppose that $x_1>f(x_1)$ for some $x_1.$ Then $(x,y)=(x_1,\frac{1}{x_1-f(x_1)})$ gives us $$x_1=f(x_1)-\frac{1}{f(y)}<f(x_1),$$contradiction. So $\boxed{x \le f(x) \hspace{2mm} \forall x}.$ But this means that $$x+\frac{1}{f(y)}=f(\frac{1}{y}+f(x))\ge \frac{1}{y}+f(x)\ge \frac{1}{y}+x \Longrightarrow$$$$\frac{1}{f(y)}\ge \frac{1}{y}\Longrightarrow \boxed{y\ge f(y) \hspace{2mm} \forall y}.$$Combining the two boxed results gives us $f(x)=x.$

Amin12 wrote: Find all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that $$\frac{x+f(y)}{xf(y)}=f\left(\frac{1}{y}+f\left(\frac{1}{x}\right)\right)$$for all positive real numbers $x$ and $y$. Equivalently, $$\frac{1}{x}+\frac{1}{f(y)}=f\left(\frac{1}{y}+f\left(\frac{1}{x}\right)\right)$$for all $x,y>0$. Put $t=\tfrac{1}{x}$ thus $t+\tfrac{1}{f(y)}=f\left(f(t)+\tfrac{1}{y}\right)$ for all $t,y>0$. Observe that as $t \rightarrow \infty$ we see $\mathbb{R}(f)$ has no upper bound. Thus, for any $\varepsilon>0$ it is possible to pick $y$ with $\tfrac{1}{f(y)}<\varepsilon$; hence $ \cup [\tfrac{1}{f(y)}, \infty)$ is a subset of $\mathbb{R}(f)$, consequently $f$ is surjective over $\mathbb{R}^{+}$. If $f(a)=f(b)$ then plugging $t=a$ and $t=b$ subsequently, we conclude $a=b$ or $f$ is injective. Thus, $f$ is a bijection on positive reals. Now substitute $y=\tfrac{1}{f(z)}$ yielding $$f(f(t)+f(z))=t+\frac{1}{f\left(\frac{1}{f(z)}\right)}$$for all $t,z>0$. Swapping $t,z$ fixes the LHS, hence $\frac{1}{f\left(\frac{1}{f(z)}\right)}=z+C$ for some constant $C$ and all $z>0$. Hence $f(f(t)+f(z))=t+z+C$ for all $t,z>0$. Playing the Devil's trick again, we put $x=f(z)$ in the original equation; so $$\frac{1}{f(z)}+\frac{1}{f(y)}=f\left(\frac{1}{y}+\frac{1}{z+C}\right)$$and swap $y,z$; injectivity of $f$ then yields that $y \mapsto \frac{1}{y}-\frac{1}{y+C}$ is a constant function. Thus, $C=0$ and so $f(f(y)+f(z))=y+z$ for all $y,z>0$. Now plug $x \mapsto f\left(\frac{1}{x}\right)$ in $f\left(\frac{1}{f(x)}\right)=\frac{1}{x}$ to conclude that $f\left(\frac{1}{x}\right)=\frac{1}{f(x)}$ for all $x>0$. Now we immediately get $f(f(x))=x$ for all $x>0$ and so $f(a+b)=f(a)+f(b)$ (putting $a=f(y), b=f(z)$); hence $f$ is additive too. Now $f$ is strictly increasing so if $f(x_0)>x_0$ then $x_0=f(f(x_0)>x_0$ and vice-versa. Thus, $f(x_0)=x_0$ for all $x_0>0$ and $f$ is the identity function. It also works. $\blacksquare$

Fix $y$ to note that $f$ is injective and $(\frac{1}{f(y)},\infty) \subseteq f$. Now take $y$ with $f(y) \to \infty$ gives $(0,\infty) = f$. Therefore $f$ is bijective. Now note that $$f(x+\frac{1}{f(y)} + \frac1z) = f(f(f(x) + 1/y ) + 1/z) = f(x) + 1/y + 1/f(z)$$Replace $z$ by $1/z$ in the above relation and consider symmetric equation with $x,z$ swapped we deduce $$f(x) = 1/f(1/x) + \underbrace{f(z) - 1/f(1/z)}_c$$Fix $z$ and vary $x$ to get using $f$ bijective that $(c,\infty) = f$ so $c = 0$, i.e. $f(z) = \frac 1 {f(1/z)}$. Now we rewrite original FE as $$f(f(x) + y) = x + f(y).$$Replace $x$ by $f(x)$ to get $$f(f^2(x) + y) = f(x) + f(y) = f(y) + f(x) = f(f^2(y) + f(x)). \qquad\qquad \dots \dots (1)$$Now use injectivity to get $f^2(x) = \underbrace{f^2(y) - y}_{c'} + x$. Taking a similar consideration as before with $c$ we see that $c' = 0$. Therefore $f(x)^2 = x$, so $(1)$ becomes Cauchy FE. Extend $f(x)$ by $f(-x) = -f(x)$ for all $x < 0$ and note that extended $f$ satisfies Cauchy on the reals, and $f(x) > 0$ for all $x > 0$, so $f$ is linear, from which we conclude that $f(x) = x$, as desired.

Enjoyed this a whole lot Amin12 wrote: Find all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that $$\frac{x+f(y)}{xf(y)}=f\left(\frac{1}{y}+f\left(\frac{1}{x}\right)\right)$$for all positive real numbers $x$ and $y$. We begin our solution by claiming that $\boxed{f(x) \equiv x \forall \ x \in \mathbb{R^+}}$ is the only solution to the given equation. Note that it indeed works. $ $ Now, let $P(x,y)$ denote the given assertion. $(\star) P \left(\frac{1}{x}, \frac{1}{f(y)}\right) : $ \begin{align*} x + \frac{1}{f(\frac{1}{f(y)})} & = f(f(y) + f(x)) \\ & = f(f(x) + f(y)) \\ & = y + \frac{1}{f (\frac{1}{f(x)})} \\ \end{align*}Therefore, let $h(x) =\frac{1}{f \left(\frac{1}{f(x)}\right)} $. Then we have $h(x) - h(y) = x - y \implies h(x) = x + c \forall x \in \mathbb{R^+}$ and some $c \in \mathbb{R}$. So, $f(\frac{1}{f(y)}) = \frac{1}{y+c} \implies f$ is injective . Now $P(f(x) , y) ; P(f(y),x)$ imply that $c = 0$. So $f(\frac{1}{f(y)}) = \frac{1}{y}$ and hence $f$ is bijective . Thus, $P(f(x),y ) \implies \frac{1}{f(x)} + \frac{1}{f(y)} = f\left(\frac{1}{x} + \frac{1}{y} \right)$. $ $ Call the last equation $Q\left(\frac{1}{x} , \frac{1}{y}\right)$. Then, $Q(f(x) , f(y)) : f(f(x) + f(y)) = x + y \forall x, y \in \mathbb{R^+}$. $(\star \star) Q(x,x) : f(2f(x)) = 2x$ $ $ $(\star \star \star) Q(2f(x) , 2f(y)): f(2x + 2y ) = 2\cdot (f(x) + f(y))$. (By induction on this, we get $f(2^nx + 2^ny) = 2^nf(x) +2^nf(y)$ In this equation replace $x \mapsto (x+z)$ and observe that: $$ \underbrace{f(x+z) + f(y) = \frac{1}{2} \cdot \left(f( 2x + 2y + 2z)\right) = f(x+y) + f(z)}_{S(x,y,z)}$$Now using $S(x,x, 3x) : f(2x) + f(3x) = 5 \cdot f(x)$ and $S(x,2x,3x) : 2 \cdot f(3x) - f(2x) = 4 f(x)$. $ $ Combining both these equations, we get that $f(2x) = 2f(x)$. So , iterating $f$ on both sides we get $f(f(2x)) = f(2f(x)) = 2x \implies f(f(x)) = x \forall x \in \mathbb{R^+}$. $ $ So, $f$ is an involution. Now comparing $Q(x,y)$ and $f(f(x+y))$ we see that $$f(f(x)+f(y)) = x + y = f(f(x+y)) \overset{\text{injection}}{\implies} \underbrace{f(x) + f(y) = f(x+y)}_{\text{Cauchy}} \implies f \equiv \alpha x + \beta $$Plugging this back in $P(x,y)$ we get $\beta = 0$ and $\alpha = 1$.

Denote the assertion as $P(x,y)$. As $x\to 0$, we get unbounded values of $f$, and for a fixed value of $y$, we get that $f(x)$ is surjective over $\left(\frac{1}{f(y)},\infty\right)$. So, letting $f(y)$ tend towards infinity gives surjectivity. If $f(x_1)=f(x_2)$, $P\left(\frac{1}{x_1},y\right),P\left(\frac{1}{x_2},y\right)$ gives $x_1=x_2$, so $f$ is bijective. Consider $P\left(x,\frac{1}{y}\right)$. As $x$ varies for fixed $y$, we get that the image of $(y,\infty)$ is $\left(\frac{1}{f(1/y)},\infty\right)$. Thus, if $y_1<y_2$, $\frac{1}{f(1/y_1)}<\frac{1}{f(1/y_2)}$, so $f$ is increasing. As it is both increasing and bijective, our function must be continuous. Consider $P\left(\frac{1}{y-\frac{1}{f(x)}},x\right)$. This gives that $f\left(\frac{1}{x}+f\left(y-\frac{1}{f(x)}\right)\right)=y-\frac{1}{f(x)}+\frac{1}{f(x)}=y$, so it is fixed as $x$ varies. Note that we must have $x>f^{-1}\left(\frac{1}{y}\right)$ to make sure the argument is positive, and as $x$ approaches this lower bound, the LHS gets arbitrarily close to $C=f\left(\frac{1}{f^{-1}\left(1/y\right)}\right)$. The RHS cannot be less than $C$, since it will otherwise be exceeded as $x$ approaches $f^{-1}\left(\frac{1}{y}\right)$. Likewise, it can't be more than $C$. So, we must have $f\left(\frac{1}{x}+f\left(y-\frac{1}{f(x)}\right)\right)=C=f\left(\frac{1}{f^{-1}(1/y)}\right)\implies \frac{1}{x}+f\left(y-\frac{1}{f(x)}\right)=\frac{1}{f^{-1}(1/y)}$. As $x\to\infty$, LHS approaches $f(y)$, so we can get by similar logic that it must always be $f(y)$. Thus, $f^{-1}(1/y)=\frac{1}{f(y)}\implies f\left(\frac{1}{f(y)}\right)=\frac{1}{y}$. Finally, consider $P(x,f(y))$, which gives $\frac{1}{x}+\frac{1}{f(f(y))}=f\left(\frac{1}{f(y)}+f\left(\frac{1}{x}\right)\right)$. As $x\to\infty$, LHS approaches $\frac{1}{f(f(y))}$ while RHS is always larger, but approaches $\frac{1}{y}$ by above. Using a similar argument, if $\frac{1}{y}>\frac{1}{f(f(y))}$, then we eventualy have RHS>LHS, and if $\frac{1}{f(f(y))}>\frac{1}{y}$, we have the opposite. Hence, $\frac{1}{f(f(y))}=\frac{1}{y}\implies f(f(y))=y$. Now, we have that $f$ is both an involution and increasing. If it is nonconstant, though, we can find $a<b$ such that $f(a)=b>f(b)=a$. Thus, $f(x)=x$ is the only solution.

Amin12 wrote: Find all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that $$\frac{x+f(y)}{xf(y)}=f(\frac{1}{y}+f(\frac{1}{x}))$$for all positive real numbers $x$ and $y$. This is Hard(and Nice) Claim 1.$f$ is injective function. Proof: $$P(\frac{1}{x},\frac{1}{f(y)})\implies f(f(x)+f(y))=x+\frac{1}{f(\frac{1}{f(y)})}=y+\frac{1}{f(\frac{1}{f(x)})}\implies f(\frac{1}{f(x)})=\frac{1}{x+c}$$$f(\frac{1}{f(x)})=\frac{1}{x+c}\implies f\to injective$ Claim 2.$f(\frac{1}{f(x)})=\frac{1}{x}$ Proof: $P(f(x),y)\implies \frac{1}{f(x)}+\frac{1}{f(y)}=f(\frac{1}{y}+f(\frac{1}{f(x)}))=f(\frac{1}{y}+\frac{1}{x+c})$ $P(f(y),x)\implies \frac{1}{f(y)}+\frac{1}{f(x)}=f(\frac{1}{x}+f(\frac{1}{f(y)}))=f(\frac{1}{x}+\frac{1}{y+c})$ Then $f(\frac{1}{y}+\frac{1}{x+c})=f(\frac{1}{x}+\frac{1}{y+c})\implies \frac{1}{y}+\frac{1}{x+c}=\frac{1}{x}+\frac{1}{y+c}$ Then $c=0\implies f(\frac{1}{f(x)})=\frac{1}{x}$ $P(f(x),y)\implies f(\frac{1}{x}+\frac{1}{y})=\frac{1}{f(x)}+\frac{1}{f(y)}$ $x=\frac{kt}{k+t}\implies f(\frac{1}{k}+\frac{1}{t}+\frac{1}{y})=\frac{1}{f(y)}+\frac{1}{f(\frac{kt}{k+t})}$ Then $\frac{1}{f(y)}+\frac{1}{f(\frac{kt}{k+t})}=\frac{1}{f(k)}+\frac{1}{f(\frac{yt}{y+t})}$ $\frac{1}{f(\frac{1}{x})}=g(x)$ Then $g(\frac{1}{y})+g(\frac{1}{k}+\frac{1}{t})=g(\frac{1}{k})+g(\frac{1}{y}+\frac{1}{t})$ $\frac{1}{y}\to y,\frac{1}{k}\to k,\frac{1}{t}\to t$ Then $g(y+t)-g(y)=g(k+t)-g(k)\implies g-additive$ And $g:\mathbb{R^+}\rightarrow\mathbb{R^+},additive\implies g(x)=kx+b$ $f(\frac{1}{f(x)})=\frac{1}{x}\implies g(\frac{1}{g(x)})=\frac{1}{x}\implies \frac{k}{kx+b}+b=\frac{1}{x}$ Then $b=0,k=1$ $g(x)=x,x\in R^+\implies f(x)=x,x\in R^+$

Let $P(x,y)$ the given assertion. $P(1/x,1/f(y)): 1/f(1/f(y)))+x=f(f(y)+f(x))$ $y\leftrightarrow x $: $1/f(1/f(y)))+x=1/f(1/f(x)))+y$ so $f(1/f(x))=1/(x+C)$, so $f$ is injective. Now we have that $P(1/x,1/f(y)) : x+y+C=f(f(x)+f(y)) :Q(x,y)$ $Q(x+z,y): x+y+z+C=f(f(x+z)+f(y))$ $Q(x,y+z): x+y+z+C=f(f(x)+f(y+z))$. Since $f$ is injective we have that $f(x+z)+f(y)=f(x)+f(y+z)$. Therefore $x+y+C+f(f(z))=f(f(x)+f(y))+f(f(z))=f(f(x))+f(f(y)+f(z))=f(f(x))+y+z+C$ so $f(f(x))=x+D$. $P(1/f(x),1/y): f(x)+1/f(1/y))=f(x+y+D)=f(y)+1/f(1/x))$ so $f(x)-1/f(1/x))=E$. setting $x\leftrightarrow 1/x$ it's simple to show that $E=0$ so $f(x)f(1/x)=1$. Since $f(1/f(x))=1/(x+C)$ we have that $f(f(x))=x+C$ so $C=D$. Also $1/x+D=f(f(1/x))=f(1/f(x))=1/f(f(x))=1/(x+D)\Leftrightarrow...\Leftrightarrow D=0$, thus $C=D=0$. So $f(f(x))=x$ and $f(f(x)+f(y))=x+y$. $x\rightarrow f(x),y\rightarrow f(y): f(x+y)=f(x)+f(y)$ etc.

Answer: $f(x)=\frac{1}{x} \ \ \forall x\in \mathbb{R}^+.$ Proof: It's easy to see that the above function is a solution. Let $P(x,y)$ denote the given assertion, by comparing \[P\left(\frac{1}{x}, \frac{1}{f(1)}\right) \implies \frac{1+xf\left(\frac{1}{f(1)}\right)}{f\left(\frac{1}{f(1)}\right)}=f(f(1)+f(x))\]and \[P\left(1,\frac{1}{f(x)}\right)\implies \frac{1+f\left(\frac{1}{f(x)}\right)}{f\left(\frac{1}{f(x)}\right)}=f(f(x)+f(1))\]we will get $f\left(\frac{1}{f(x)}\right)=\frac{1}{x+C_1}$ where $C_1=\frac{1}{f\left(\frac{1}{f(1)}\right)}-1.$ From here, it's also clear that $f$ is injective. Also, by comparing \[P(f(x),1) \implies \frac{f(x)+f(1)}{f(x)f(1)}=f\left(1+f\left(\frac{1}{f(x)}\right)\right)\]and \[P(f(1),x) \implies \frac{f(1)+f(x)}{f(1)f(x)}=f\left(\frac{1}{x}+f\left(\frac{1}{f(1)}\right)\right)\]we will get $f\left(1+f\left(\frac{1}{f(x)}\right)\right)=f\left(\frac{1}{x}+f\left(\frac{1}{f(1)}\right)\right) \implies f\left(\frac{1}{f(x)}\right)=\frac{1}{x}+C_2$ where $C_2=f\left(\frac{1}{f(1)}\right)-1.$ Therefore, we have $\frac{1}{x+C_1}=\frac{1}{x}+C_2$ or equivalently, $C_2x^2+C_1C_2x+C_1=0.$ Since this holds for all positive real $x$, it must be the case where $C_1=C_2=0$ which implies $f\left(\frac{1}{f(x)}\right)=\frac{1}{x}.$ Now notice that \[P\left(\frac{1}{x}, \frac{1}{f(x)}\right) \implies f(2f(x))=2x\]and so $4f(x)=f(2f(2f(x)))=f(4x).$ Then, \[P\left(\frac{1}{x}, \frac{1}{f(3x)}\right)\implies f(f(3x)+f(x))=4x=f(2f(2x)) \implies f(3x)+f(x)=2f(2x)\]and \[P\left(\frac{1}{2x}, \frac{1}{f(4x)}\right)\implies f(4f(x)+f(2x))=f(f(4x)+f(2x))=6x=f(2f(3x)) \implies 4f(x)+f(2x)=2f(3x)\]give us \[4f(2x)-2f(x)=2f(3x)=4f(x)+f(2x) \implies f(2x)=2f(x) \ \ \forall x\in \mathbb{R}^{+}.\]Hence, $2f(f(x))=f(2f(x))=2x \implies f(f(x))=x$ and $f\left(\frac{1}{x}\right)=f\left(\frac{1}{f(f(x))}\right)=\frac{1}{f(x)}.$ Lastly, \[P\left(f\left(\frac{1}{x}\right), \frac{1}{y}\right)\implies f(x+y)=\frac{f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)}{f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)}=f(x)+f(y)\]implies $f$ is additive over the positive reals, thus $f$ is linear and letting $f(x)=ax+b$ where $a,b$ are constants, we see that since $f(f(x))=x \implies a^2x+ab+b=x \implies a=1, b=0$, $\boxed{f(x)=x}$ is the only solution. $\quad \blacksquare$

$\frac{x+f(y)}{xf(y)} = \frac{1}{f(y)} + \frac{1}{x}$ Now put $\frac{1}{x}$ instead of $x$ in equation. Let $P(x,y) : \frac{1}{f(y)} + x = f(\frac{1}{y} + f(x))$. Let $f(a) = f(b)$ then $P(a,y) , P(b,y)$ implies that $f$ is injective. $P(x,\frac{1}{f(y)}) : \frac{1}{f(\frac{1}{f(y)})} + x = f(f(y) + f(x)) = \frac{1}{f(\frac{1}{f(x)})} + y \implies \frac{1}{f(\frac{1}{f(x)})} - x = t \implies f(\frac{1}{f(x)}) = \frac{1}{x+t}$ $P(f(x),y) : f(\frac{1}{y} + \frac{1}{x+t}) = \frac{1}{f(y)} + \frac{1}{f(x)} = f(\frac{1}{x} + \frac{1}{y+t}) \implies \frac{1}{y} + \frac{1}{x+t} = \frac{1}{x} + \frac{1}{y+t} \implies t = 0 \implies f(\frac{1}{f(x)}) = \frac{1}{x}$ $P(x,\frac{1}{f(y)}) : \frac{1}{f(\frac{1}{f(y)})} + x = f(f(y) + f(x)) \implies f(f(y) + f(x)) = x+y$ Let $f(f(y) + f(x)) = x+y$ be $Q(x,y)$. $Q(x,x) , Q(x-t,x+t) : f(x+t) - f(x) = f(x) - f(x-t)$ which holds for any $x > t > 0$ which implies that $f$ is linear so $f(x)= ax + b$. we had $f(f(y) + f(x)) = x+y \implies f(a(x+y)+2b) = x+y \implies a^2(x+y) + 2ab + b = x+y \implies (a^2-1)(x+y) + 2ab + b = 0$ which since $a,b$ are constant but $x,y$ are not implies that $a^2-1 = 0 \implies a = 1$ so $x+y + 2b + b = x+y \implies b = 0$ so $f(x) = ax + b = x$ which clearly works.

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Man this took really long Note that the above is equivalent to $\frac{1}{f(y)}+\frac{1}{x}=f(\frac{1}{y}+f(\frac{1}{x}))$, which also implies injective as a result Let $P(x,y) : \frac{1}{f(y)}+\frac{1}{x}=f(\frac{1}{y}+f(\frac{1}{x}))$ Setting $P(f(x),y)=P(f(y),x)$, we get $\frac{1}{y}-f(\frac{1}{f(y)})=\frac{1}{x}-f(\frac{1}{f(x)})$ This implies that $\frac{1}{x}-f(\frac{1}{f(x)})=-c$ for some constant $c$ over all values of $x$ Now, setting $P(x,\frac{1}{f(\frac{1}{y})})=(y,\frac{1}{f(\frac{1}{x})})$ $\frac{1}{x}-\frac{1}{x+c}=\frac{1}{y}-\frac{1}{y+c}$ Note this only holds iff $c=0$ Taking $P(x,\frac{1}{f(y)})$ $y+\frac{1}{x}=f(f(y)+f(\frac{1}{x}))$ $y+x=f(f(y)+f(x))$ Now note, if $a+b=c+d$ $f(f(a)+f(b))=a+b=c+d=f(f(c)+f(d))$ Due to injectivity, this implies $f(a)+f(b)=f(c)+f(d)$ Note $2f(3x)=f(4x)+f(2x)=3f(2x)$ Also, $5f(2x)=2f(3x)+2f(2x)=2f(4x)+2f(x)=4f(2x)+2f(x)$, so $f(2x)=2f(x)$ Recall $P(y,f(y))$ implies $\frac{f(y)}{2}f(\frac{2}{y})=1$ However, together with $f(2x)=2f(x)$, we have $\frac{1}{f(x)}=f(\frac{1}{x})$ Therefore, we have that $f(f(x))=x$ Also,$ P(\frac{1}{x},\frac{1}{y})$ gives us $f(y)+x=f(y+f(x))$ , Since $f$ is surjective, varying the value of $f(x)$, we have that $f$ is a strictly increasing function FTSOC suppose $f(x)=a$ where $a>x$ Then, note $f(a)=x<a<f(x)$ contradiction. The same holds for when $a<x$ Hence, $f(x)=x$ for all values of $x$

Amin12 wrote: Find all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that $$\frac{x+f(y)}{xf(y)}=f(\frac{1}{y}+f(\frac{1}{x}))$$for all positive real numbers $x$ and $y$. My 400th post. Let $P(x,y)$ be the assertion of the given FE, which is $$f\left(\frac1{y}+f\left(\frac1{x}\right)\right)=\frac1{f(y)}+\frac1{x},\quad\forall x,y>0.$$Assume that $f(a)=f(b)$, then $P(1/a,y)$ and $P(1/b,y)$ give $a=b$ or $f$ is injective. Hence comparing $P(f(x),y)$ and $P(f(y),x)$, we easily get $$\frac1{y}+f\left(\frac1{f(x)}\right)=\frac1{x}+f\left(\frac1{f(y)}\right)\implies f\left(\frac1{f(x)}\right)=\frac1{x}+c.$$Similarly, comparing $P(1/x,1/f(y))$ and $P(1/y,1/f(x))$, we get $$f\left(\frac1{f(x)}\right)=\frac1{x+d}\implies\frac1{x}+c=\frac1{x+d},$$for all $x>0$. Let $x\to\infty$, we get $c=d=0$, so $f\left(\frac1{f(x)}\right)=\frac1{x}$ and $P(1/x,1/f(y))$ gives $$Q(x,y):f(f(x)+f(y))=x+y,\quad\forall x,y>0.$$$Q(f(x)+f(y),y)$ gives $$f(x+y+f(y))=f(x)+y+f(y).$$Plugging $x$ by $x+f(x)$ into the above FE and changing the role of $x,y$, by the injectivity, we get $$f(x+f(x))=x+f(x)+d.$$If $d>0$, $Q(x+f(x),d)$ and the injectivity give $d+f(d)=0$, absurb. Thus $d=0$, so $Q(x,f(x))$ gives $$f(f(x)+f(f(x)))=x+f(x)=f(x+f(x))\implies f(f(x))=x.$$$Q(f(x),f(y))$ immediately implies $f$ is additive, so after checking, we get $f(x)=x$ for all $x>0$ is a solution.

Rewrite the assertion as $\frac 1x + \frac{1}{f(y)} = f(\frac 1y + f(\frac 1x))$. Claim: $f$ is bijective. Proof: For injectivity, we can vary $\frac 1x$. For surjectivity, observe we can hit any value above $\frac{1}{f(y)}$, since we can make $\frac{1}{f(y)}$ as small as we want we are done. Now let $\frac 1y = a$, we have $\frac{1}{f(\frac 1a)} < f(a + k)$ as $k$ is a positive real. Then we have $\frac 1x + \frac{1}{f(y)} = f(\frac 1y + f(\frac 1x)) > \frac{1}{f(\frac{1}{f(x)})}$, so letting $\frac{1}{f(y)}$ approach zero gives $f(\frac{1}{f(x)}) \ge \frac 1x$. Now substitute $x = f(k)$, so we have $\frac{1}{f(k)} + \frac{1}{f(y)} = f(f(\frac{1}{f(k)}) + \frac 1y ) = f(\frac 1k + \frac 1y + c)$ for $c = f(\frac{1}{f(k)}) - \frac 1k \ge 0)$. Now observe each pair $(k,y)$ results in exactly one value of $c$, which is the same as the value of $c$ given by $(y,k)$, but also this value is uniquely determined by $k$ so $c$ is constant, thus $f(\frac{1}{f(x)}) = \frac 1x + c$, but since the left hand side gets as small as we want we must have $c = 0$ and $\frac{1}{f(x)}+ \frac{1}{f(y)}= f(\frac 1x + \frac 1y)$. Now let $f(a)= 1, f(\frac{1}{f(a)} ) = f(1) = a$, then we have $f(\frac{1}{f(1)}) =f(\frac 1a) = 1$, so $a = \frac 1a$ giving $a = 1$. Now we prove that $f$ is the identity. Assume $a < b$ but $f(a) \ge f(b)$. Now we have $\frac{1}{f(a)} \le\frac{1}{f(b)} < f(\frac 1b + k) = f(\frac 1a)$, giving $f(a)f(\frac 1a) > 1$. Clearly, $a$ cannot be $1$, so $f(x) > 1$ for $x > 1$. Likewise, assume $\frac{1}{f(x)} > 1 $, which gives $\frac 1x > 1$, so $f(x) > 1$ iff $x >1$. This fact carries the rest of the solution, we can now proceed with a standard rational extension. First we solve $f$ over rationals. Let $Q$ be the assertion $\frac{1}{f(x)} + \frac{1}{f(y)} = f(\frac 1x + \frac 1y)$. Now $Q(1,1)$ gives $f(2) = 2$, then we can use the assertion $R$, being $f(\frac{1}{f(x)}) = \frac 1x$ to get $f(\frac 12) = 2$. Now we can always induct, do $Q(1, \frac 1n)$ to get $f(n + 1) =n + 1$ and $R(n + 1)$ to get $f(\frac{1}{n + 1}) = \frac{1}{n +1}$. Now to get all rationals we induct on the denominator, we can use $Q(\frac 1n, 2)$ to get all rationals with denominator $2$, then use $Q(\frac 1n, 3)$ and $Q(\frac 1n, \frac 32)$ to get all rationals with denominator $3$, and so one and so forth we win. To finish, we prove $f(r) = r$ for all reals. First take $r > 1$. Assume $f(r) > r$. Then there exists some rational $q$ with $\frac{1}{f(q)} + \frac{1}{f(r)} < 1 < \frac 1q + \frac 1r$, contradiction. Symmetrical argument proves $f(r) = r$ for $r > 1$. Now consider $R(\frac 1r)$, this gives $f(\frac{1}{f(\frac 1r)}) = r$, giving $\frac 1r = f(\frac 1r)$ by injectivity.