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Source: Iran 3rd round 2017 first Algebra exam

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Amin12

#1 h Aug 7, 2017, 8:35 AM • 3 Y

Y by yayitsme, Adventure10, Mango247

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 post has been edited 2 times. Last edited by Amin12, Aug 7, 2017, 8:40 AM

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Vietnamisalwaysinmyheart

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Vietnamisalwaysinmyheart

#2 h Aug 7, 2017, 3:04 PM • 4 Y

Y by gemcl, Jonathankirk, Adventure10, Mango247

Here is my solution:
Click to reveal hidden text

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#3 h Aug 7, 2017, 3:07 PM • 3 Y

Y by Adventure10, Mango247, math_comb01

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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TLP.39

#4 h Sep 28, 2017, 10:26 AM • 2 Y

Y by Adventure10, Mango247

Another solution.

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#5 h Feb 20, 2018, 12:10 PM • 23 Y

Y by TLP.39, Ankoganit, k.vasilev, Xurshid.Turgunboyev, naw.ngs, gemcl, MahdiTA, Kayak, rmtf1111, e_plus_pi, Sillyguy, ValidName, Aryan-23, Arefe, r_ef, maryam2002, Gaussian_cyber, FAA2533, electrovector, Kamikaze-1, Adventure10, TemetNosce, NicoN9

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 $f(x)=x.$

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#6 h Apr 8, 2018, 8:28 PM • 3 Y

Y by e_plus_pi, Adventure10, Mango247

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. $\blacksquare$

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#8 h May 2, 2019, 3:17 PM • 2 Y

Y by Adventure10, Mango247

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.

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#9 h May 4, 2019, 7:59 AM • 1 Y

Y by Adventure10

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.
$\text{[math]}$

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, $P(f(x),y ) \implies \frac{1}{f(x)} + \frac{1}{f(y)} = f\left(\frac{1}{x} + \frac{1}{y} \right)$ .
$\text{[math]}$
Call the last equation $Q\left(\frac{1}{x} , \frac{1}{y}\right)$ . Then, $Q(f(x) , f(y)) : f(f(x) + f(y)) = x + y \forall x, y \in \mathbb{R^+}$ .
$(\star \star) Q(x,x) : f(2f(x)) = 2x$
$\text{[math]}$
$(\star \star \star) Q(2f(x) , 2f(y)): f(2x + 2y ) = 2\cdot (f(x) + f(y))$ . (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 $S(x,2x,3x) : 2 \cdot f(3x) - f(2x) = 4 f(x)$ .
$\text{[math]}$
Combining both these equations, we get that $f(2x) = 2f(x)$ . So , iterating $f$ on both sides we get $f(f(2x)) = f(2f(x)) = 2x \implies f(f(x)) = x \forall x \in \mathbb{R^+}$ .
$\text{[math]}$
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 $\alpha = 1$ .

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#11 h Dec 24, 2019, 3:24 AM • 1 Y

Y by Adventure10

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.

This post has been edited 2 times. Last edited by william122, Dec 24, 2019, 12:33 PM

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Functional_equation

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Functional_equation

#12 h Jan 5, 2021, 3:45 PM • 1 Y

Y by amar_04

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 $b=0,k=1$
$g(x)=x,x\in R^+\implies f(x)=x,x\in R^+$

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Prod55

#13 h Jun 12, 2021, 1:39 PM

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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.

This post has been edited 1 time. Last edited by Prod55, Jun 12, 2021, 1:54 PM
Reason: typo

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#14 h Jul 7, 2021, 5:19 AM

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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. $\quad \blacksquare$

This post has been edited 2 times. Last edited by Keith50, Jul 7, 2021, 5:21 AM

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#15 h Jul 28, 2022, 10:17 AM

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$\frac{x+f(y)}{xf(y)} = \frac{1}{f(y)} + \frac{1}{x}$ 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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ZETA_in_olympiad

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ZETA_in_olympiad

#16 h Aug 11, 2022, 8:36 PM • 1 Y

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Also see here.

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#17 h Apr 17, 2023, 3:21 PM

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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 $x$

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ThisNameIsNotAvailable

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ThisNameIsNotAvailable

#18 h May 10, 2023, 1:51 PM

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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.

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ezpotd

#19 h Feb 9, 2024, 9:49 PM

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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.

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#21 h Apr 30, 2025, 11:46 AM

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$\frac{1}{f(y)}+x=f(\frac{1}{y}+f(x))$ Answer is $f(x)=x$ which holds. Let $P(x,y)$ be the assertion.
Claim: $f$ is bijective.
Proof: If $f(a)=f(b)$ , compare $P(a,y)$ and $P(b,y)$ to get contradiction. Fix $y$ and increase $x$ to see that $f$ takes all sufficiently large positive values. We can pick $1/f(y)$ sufficiently small thus, $f$ is surjective.
Claim: $f(\frac{1}{f(x)})=\frac{1}{x}$ .
Proof: Plug $P(x,1/f(y))$ in order to observe that
$\frac{1}{f(\frac{1}{f(y)})}+x=f(f(x)+f(y))=\frac{1}{f(\frac{1}{f(x)})}+y\implies x-\frac{1}{f(\frac{1}{f(x)})}=y-\frac{1}{f(\frac{1}{f(y)})}$ Since this implies $x-1/f(1/f(x))$ is constant and it's smaller than any positive $y$ , it must be a nonpositive constant. Let $\frac{1}{f(\frac{1}{f(x)})}=x+c$ or $f(\frac{1}{f(x)})=\frac{1}{x+c}$ where $c\geq 0$ . By symmetry and injectivity we have
$f(\frac{1}{x}+\frac{1}{y+c})=f(\frac{1}{x}+f(\frac{1}{f(y)}))=\frac{1}{f(y)}+\frac{1}{f(x)}=f(\frac{1}{y}+f(\frac{1}{f(x)}))=f(\frac{1}{y}+\frac{1}{x+c})$ Thus, $\frac{1}{x}-\frac{1}{x+c}=\frac{c}{x(x+c)}$ is constant which requires $c=0$ . So $f(1/f(x))=1/x$ as we have claimed.
Claim: $f$ is involution and $f(x)f(\frac{1}{x})=1$ .
Proof: $P(x,1/f(y))$ gives $x+y=f(f(x)+f(y))$ . We get $f(f(x)+f(y+z))=x+y+z=f(f(x+z)+f(y))$ and injectivity implies $f(x+y)-f(x)$ is independent of $x$ . Let $f(x+y)-f(x)=h(y)$ . Since $h(x)+f(y)=f(x+y)=h(y)+f(x)$ we observe $h(x)=f(x)+d$ . Thus, $f(x+y)=f(x)+f(y)+d$ .
$f(f(x))+f(\frac{1}{y})+d=f(f(x)+\frac{1}{y})=x+\frac{1}{f(y)}$ $f(f(x))-x=t$ and $f(\frac{1}{y})-\frac{1}{f(y)}=r$ . Since $r=f(\frac{1}{f(x)})-\frac{1}{f(f(x))}=\frac{1}{x}-\frac{1}{x+t}=\frac{t}{x(x+t)}$ , we must have $t=0$ which implies $r=0$ . Thus, $f$ is an involution and $f(x)f(\frac{1}{x})=1$ .
Claim: $f$ is additive.
Proof: $P(f(x),1/y)$ yields $f(x)+f(y)=f(x+y)$ .

Since $f$ is additive and $f$ takes values on positive reals, $f$ is Cauchy function hence $f(x)=cx$ which implies $c=1$ . So $f(x)=x$ is the only solution as desired. $\blacksquare$

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