Find all $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all distinct $x,y,z$ $f(x)^2-f(y)f(z)=f(x^y)f(y)f(z)[f(y^z)-f(z^x)]$
Problem
Source: Pakistan TST(2). P3. Proposed by Awais Chisti and Sualeh Asif
Tags: functional equation, Functional Equations, algebra proposed, algebra, algebraic identities
mihajlon
29.01.2017 00:03
Let $A(x,y,z)=f(x^2)-f(y)f(z)=f(x^y)f(y)f(z)[f(y^z)-f(z^x)]$ be given assertion.
$$A(1,1,1)\Longrightarrow f(1)-f(1)^2=f(1)^3(f(1)-f(1))=0\Longleftrightarrow f(1)=1$$.
Now \begin{align*}
A(1,y,z)&\Longrightarrow 1-f(y)f(z)=f(y)f(z)(f(y^z)-f(z))\\ & \Longleftrightarrow 1=f(y)f(z)(f(y^z)-f(z)+1)=S
\end{align*}.
Similarly from \begin{align*}
A(1,z,y)&\Longrightarrow f(y)f(z)(f(z^y)-f(y)+1)=1=S=f(y)f(z)(f(y^z)-f(z)+1)
\\ & \Longleftrightarrow f(y^z)-f(z^y)=f(z)-f(y)\ . \ . \ . \ \bigstar
\end{align*}.
Now from \begin{align*}
A(y,y,z) \Longrightarrow f(y^2)-f(y)f(z)&=f(y^y)f(y)f(z)(f(y^z)-f(z^y))
\\ & \stackrel{\bigstar}{=}f(y^y)f(y)f(z)(f(z)-f(y)):=B(y,z)
\\ & \stackrel{B(y,y)}{\Longrightarrow}f(y^2)=f(y)^2\ . \ . \ . \ \bigstar \bigstar
\end{align*}\begin{align*}
B(y,z)& \Longrightarrow f(y^y)f(y)f(z)(f(z)-f(y))=f(y^2)-f(y)f(z)
\\ & \stackrel{\bigstar \bigstar}{=} f(y)(f(y)-f(z))
\\ & \stackrel{f(y)\neq f(z)}{\Longleftrightarrow} f(y^y)f(y)f(z)=-f(y)
\\ & \Longrightarrow \text{contradiction}
\\ & \Longrightarrow f(y)=f(z)
\end{align*}Thus $f$ is constant function, which obviously not satisfy $A(x,y,z)$, thus no solutions.
math90
29.01.2017 00:07
Your solution is not valid since $x,y,z$ should be distinct.
mihajlon
29.01.2017 00:09
math90 wrote: Your solution is not valid since $x,y,z$ should be distinct. oh, thanks, I totally misread the problem...
sualehasif996
29.01.2017 00:53
I have fixed a small typo..
mihajlon
29.01.2017 02:42
I hope, that now it's okay(PM me or write below, if there is something unclear...)... Let $A(x,y,z)=f(x)^2-f(y)f(z)=f(x^y)f(y)f(z)[f(y^z)-f(z^x)]$ be given assertion. From \begin{align*} A(\underbrace{x,y,1}_{x\neq y\neq 1\neq x})& \Longrightarrow f(x)^2-f(y)f(1)=f(x^y)f(y)f(1)(f(y)-f(1)):=B(x,y),\forall x\neq 1 \\ &\Longrightarrow [ f(y)> f(1)\Longleftrightarrow f(x)^2> f(y)f(1) ] \vee [f(y)< f(1)\Longleftrightarrow f(x)^2< f(y)f(1)]\ . \ . \ . \ \spadesuit \end{align*}Also note that if there exists $1\neq m\in \mathbb{R}^+$ such $f(m)=f(1)$ then from $B(x,m)\Longrightarrow f(x)=f(1), \forall x\neq 1$, and thus function is constant and this obviously satisfy condition for any $c\in \mathbb{R}^+$ i.e. $\boxed{f(x)=c}$. Thus now we know that $$f(r)=f(1)\Longleftrightarrow r=1\ . \ . \ . \ \bigstar$$Thus assume first(case 1...) that $\exists y$ such that $f(y)> f(1)\stackrel{\spadesuit}{\Longrightarrow} f(x)^2> f(y)f(1)> f(1)^2\Longleftrightarrow f(x)> f(1)$(this obviously should hold for every $x\neq y$, since $x$ doesn't depend on $y$ in equation, this inequality also holds for $f(y)>f(1)$, since that is case that we consider now ...) , but as this statement obviously holds for every $\mathbb{R}^+\ni x\neq 1$, there can not exist $y$ such that $f(y)< f(1)$, thus $f(1)$ is minimum of function $f$. . . $\clubsuit$ Now \begin{align*} A(\underbrace{x,1,z}_{x\neq 1\neq z\neq x})& \Longrightarrow f(x)^2-f(1)f(z)=f(x)f(y)f(z)(f(1)-f(z^x))\stackrel{\clubsuit \& \bigstar}{<}0 \\ & \Longrightarrow f(x)^2<f(1)f(z)\stackrel{\clubsuit}{<}f(z)^2 \\ & \Longleftrightarrow f(x)<f(z), \forall x\neq z\neq 1\neq x \end{align*}which is obviously contradiction(if you wonder why, then just assume that $f(a)>f(b)$ for $z=a$ and $x=b$, but then for $z=b$ and $x=a$ inequality doesn't holds... ) Thus assume now(case 2...) that $\exists y$ such that $f(y)< f(1)\stackrel{\spadesuit}{\Longrightarrow} f(x)^2< f(y)f(1)< f(1)^2\Longleftrightarrow f(x)< f(1)$, but as this statement obviously holds for every $\mathbb{R}^+\ni x\neq 1$, there can not exist $y$ such that $f(y)> f(1)$, thus $f(1)$ is maximum of function $f$. . . $\clubsuit \clubsuit$ Now \begin{align*} A(\underbrace{x,1,z}_{x\neq 1\neq z\neq x})& \Longrightarrow f(x)^2-f(1)f(z)=f(x)f(y)f(z)(f(1)-f(z^x))\stackrel{\clubsuit\clubsuit\& \bigstar}{>}0 \\ & \Longrightarrow f(x)^2>f(1)f(z)\stackrel{\clubsuit\clubsuit}{>}f(z)^2 \\ & \Longleftrightarrow f(x)>f(z), \forall x\neq z\neq 1\neq x \end{align*}which is obviously contradiction(if you wonder why, then just assume that $f(a)<f(b)$ for $z=a$ and $x=b$, but then for $z=b$ and $x=a$ inequality doesn't holds... )
math90
29.01.2017 08:36
My solution: Multiply each side by $f(x)$ to get $f(x)^3-f(x)f(y)f(z)=f(x)f(y)f(z)f(x^y)[f(y^z)-f(z^x)]$ Let this assertion be $P(x,y,z)$. $P(x,y,z)+P(y,z,x)+P(z,x,y)\implies f(x)^3+f(y)^3+f(z)^3=3f(x)f(y)f(z)$ for all distinct $x,y,z$. By AM-GM $f(x)^3+f(y)^3+f(z)^3\geq 3f(x)f(y)f(z)$ Since equality holds, we have $f(x)=f(y)=f(z)$ for all distinct $x,y,z$. Hence $\boxed{f(x)=c}$, where $c\in\mathbb R^+$. Clearly all these solutions work.
dangerousliri
29.01.2017 11:50
math90 wrote: My solution: Multiply each side by $f(x)$ to get $f(x)^3-f(x)f(y)f(z)=f(x)f(y)f(z)f(x^y)[f(y^z)-f(z^x)]$ Let this assertion be $P(x,y,z)$. $P(x,y,z)+P(y,z,x)+P(z,x,y)\implies f(x)^3+f(y)^3+f(z)^3=3f(x)f(y)f(z)$ for all distinct $x,y,z$. By AM-GM $f(x)^3+f(y)^3+f(z)^3\geq 3f(x)f(y)f(z)$ Since equality holds, we have $f(x)=f(y)=f(z)$ for all distinct $x,y,z$. Hence $\boxed{f(x)=c}$, where $c\in\mathbb R^+$. Clearly all these solutions work. Awesome solution .
GorgonMathDota
11.06.2019 16:57
sualehasif996 wrote: Find all $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all distinct $x,y,z$ $f(x)^2-f(y)f(z)=f(x^y)f(y)f(z)[f(y^z)-f(z^x)]$ How bout a similar functional equation but : \[ f(x^2) - f(y)f(z) = f(x^y) f(y) f(z) [f(y^z) - f(z^x) \]Cause that is what was written in the other book with the same source (?) NB : Maybe a typo but still wonder if it is still solvable
BlazingMuddy
15.06.2019 01:24
The beginning idea is pretty much the same except that now the equation is: $$ f(x)f(x^2) + f(y)f(y^2) + f(z)f(z^2) = 3f(x)f(y)f(z) $$ for all pairwise distinct positive real numbers $x$, $y$, and $z$.
Let $S$ be a subset of $\mathbb{R}$ with at least $5$ elements. Find all functions $f, g : S \rightarrow \mathbb{R}$ such that for all distinct elements $x, y, z \in S$,
$$ g(x) + g(y) + g(z) = 3f(x)f(y)f(z) $$
AnswerEither there exists 2 distinct elements $a,b\in S$ such that $f(x) = 0$ for all $x\in S\setminus\{a, b\}$ and $g\equiv 0$, or there exists real numbers $r$, $s$, $a$ (where $a\in S$; $r$ and $s$ could be equal) such that:
$$ f(x) = \begin{cases} r, & x\neq a \\ s, & x = a\end{cases} , g(x) = \begin{cases} r^3, & x\neq a \\ (3s - 2r) r^2, & x = a\end{cases} $$
SolutionObserve that this means for any 4 distinct elements $x, y, z, w\in S$,
$$ g(z) - g(w) = 3f(x)f(y) (f(z) - f(w)) \ldots (1)$$
Suppose that $f$ is non-constant. Let $c$ and $d$ be two elements in $S$ such that $f(c) \neq f(d)$. Then, the above equation reads as $f(x)f(y) = \frac{g(c) - g(d)}{3(f(c) - f(d))}$ for all $x, y\in S\setminus \{c, d\}$ with $x\neq y$. In particular, since $S\setminus \{c, d\}$ contains at least 3 elements, either $f(S\setminus \{c, d\})$ is a singleton (a set with one element) or there exists a real number $t\in S\setminus \{c, d\}$ such that $f(x) = 0$ for all $x\in S\setminus \{c, d, t\}$.
Since $|S| \geq 5$, the latter case means there exists a real number $x_1, x_2$ such that $f(x_1) = f(x_2) = 0$. This means by plugging in into the original equation,
$$ g(x_1) + g(c) + g(d) = g(x_1) + g(c) + g(t) = g(x_1) + g(d) + g(t) = g(x_1) + g(x_2) + g(t) = g(x_2) + g(c) + g(t) = 0 $$
This means $g(c) = g(d) = g(t) = g(x_2) = g(x_1) = 0$. Then $f(c) f(d) f(t) = 0$, so one of $f(c), f(d), f(t)$ turns out to be equal $0$.
Either way, there exists a real number $r$ such that there exists at most 2 real numbers $x\in S$ with $f(x) \neq r$. Note that $(1)$ means that $g(x) = g(y)$ whenever $f(x) = f(y)$, so now let $r'$ be a real number such that $g(x) = r'$ for all $x\in S\setminus \{a, b\}$. Since $|S| \geq 5$, the original equation implies $r' = r^3$.
Suppose that there exists exactly 2 such real numbers, $a$ and $b$. Then, we get:
\begin{align*}
2r^3 + g(a) &= 3r^2 f(a) \\
2r^3 + g(b) &= 3r^2 f(b) \\
r^3 + g(a) + g(b) &= 3r f(a) f(b)
\end{align*}
So, $3r^2 f(a) + 3r^2 f(b) = 3r^3 + 3r f(a) f(b)$; equivalently, $r (r - f(a))(r - f(b)) = 0$, so since $f(a), f(b)\neq r$, we get $r = 0$. It is easy to check that in this case, $g(x) = 0$ for all $x\in S$.
Otherwise, assuming $f$ is not constant, there exists only one such element, say $a$. Now, the original equation implies $2r^3 + g(a) = 3r^2 f(a)$; i.e. $g(a) = 3r^2 f(a) - 2r^3$. It is easy to check that it also satisfies the equation.
$f\equiv 1$.
Going back to the problem above, since $f$ maps $\mathbb{R}^+$ to $\mathbb{R}^+$, then it does not attain the value $0$. Let $g(x) = f(x)f(x^2)$ for all $x\in \mathbb{R}^+$. Our result above says that there exists positive real numbers $r$ and $a$ such that $f(x) = r$ and $g(x) = r^3$ whenever $x\neq a$, while $g(a) = (3f(a) - 2r)r^2$. But since $g(x) = f(x)f(x^2)$, taking a positive real number $x \not\in \{a, \sqrt{a}\}$, this implies $r^2 = r^3$ and thus $r = 1$.
Now, assume that $f(a) \neq 1$. If $a\neq 1$, then $\sqrt{a} \neq a$, so $1 = g(\sqrt{a}) = f(\sqrt{a})f(a) = f(a) \neq 1$; a contradiction. The last case is, as easily seen, $f(1) \neq 1$ and $f(x) = 1$ for all $x\neq 1$. Plugging $x = 1$, $y = 2$, and $z = 3$ into the above equation (only the $x$ matters though) yields $f(1)^2 + 2 = 3f(1)$; factoring yields $f(1) = 2$ as $f(1) \neq 1$.
Finally, we check if the function $f(x) = \begin{cases} 1, & x\neq 1 \\ 2, & x = 1\end{cases}$ satisfy the original equation... which is not the case as substituting $x = 1$, $y = 2$ and $z = 3$ yields that the left hand side is $1$ while the right hand side is $0$. So, $f\equiv 1$.
khanhnx
21.07.2019 17:44
Here is my solution for this problem Solution $f^2(x) - f(y)f(z) = f(x^y)f(y)f(z)[f(y^z) - f(z^x)]$ Let $f(1) = a$ Let $x = 1$, we have: $a^2 - f(y)f(z) = af(y)f(z)[f(y^z) - f(z)]$ $(1)$ Let $y = 1$, $z$ be $y$, $x$ be $z$, we have: $f^2(z) - af(y) = af(y)f(z)[a - f(y^z)]$ $(2)$ From $(1)$, $(2)$, we have: $a^2 - f(y)f(z) + f^2(z) - af(y) = af(y)f(z)[a - f(z)]$ $(3)$ In $(3)$, let $y$ be $z$, $z$ be $y$, we have: $a^2 - f(z)f(y) + f^2(y) - af(z) = af(z)f(y)[a - f(y)]$ $(4)$ From $(3)$, $(4)$, we have: $f^2(z) - af(y) + af(y)f^2(z) = f^2(y) - af(z) + af^2(y)f(z)$ Then: $[f(z) - f(y)][af(y)f(z) + f(y) + f(z) + a] = 0$ So: $f(y) = f(z)$ or $f(x) = c, \forall x \in \mathbb{R^+}, c \in \mathbb{R^+}$ Retry, we see that: $f(x) = c$ satisfies the problem In conclusion, we have: $f(x) = c, \forall x \in \mathbb{R^+}, c \in \mathbb{R^+}$
HasnatFarooq
23.03.2024 15:53
Let $P(x,y,z)$ be the assertion for this FE. Then $P(x,y,z)+P(y,z,x)+P(z,x,y)$ gives $f(x)^2+f(y)^2+f(z)^2=f(x)f(y)+f(y)f(z)+f(z)f(x)$. By AM-GM we have $f(x)^2+f(y)^2\ge 2f(x)f(y)$. As equality exists so we have $f(x)=f(y)=f(z)=c$ for some constant $c \in \mathbb{R}^+$.