Consider $f:\mathbb{R}^+ \to \mathbb{R}$ such that $f(3)=1$ and

$$f(x)f(y)+f(3/x) f(3/y)=2f(xy)\;\;\;\;\;\;\forall x,y \in \mathbb{R}^+$$ then choose the correct option(s):

(A) $f(2014)+f(2015)-f(2010)=100$

(B) $f$ is an even function

(C) $\frac{f(100)}{f(10)+f(90)}=\frac{1}{2}$

(D) $f$ is a periodic function

By putting $x=1$ in given equation to get $f(1)=1$ Generally I have been taught to solve such question using first principle of differentiation but here I am not able to deal with the second term of L.H.S. in given equation. May someone provide some help?


If we write $y=3/x$ then we get $$2f(x)f({3\over x}) = 2f(3)=2$$ so we have for all $x>0$: $$ f\big({3\over x}\big) = {1\over f(x)}\;\;\;(*)$$ Pluging this in original equation we get $$ f(x)f(y) +{1\over f(x)f(y)} = 2f(xy)$$ which is valid for all $x,y>0$. Put now $x=y$ we get $$ f(x)^2 +{1\over f(x)^2} = 2f(x^2)$$

By AG-GM we have: $f(x)^2 +{1\over f(x)^2}\geq 2$, so $f(x^2)\geq 1$ for all $x>0$ so $f(x)\geq 1$ for all $x>0$. Finally, useing $(*)$ we get $$ f\big({3\over x}\big) \leq 1$$ for all $x>0$. But since $x\mapsto {3\over x}$ is bijective on $\mathbb{R}^+$ we have $f(x) \leq 1$ for all $x>0$. So we conclude $f(x)=1$ for all $x> 0$, so the answer is $(C)$.

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    $\begingroup$ Double-check your inequality directions from "Finally" onwards. $\endgroup$ – J.G. Apr 17 '18 at 21:03
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    $\begingroup$ You don't need the arithmetic-geometric mean inequalities. (*) evaluated at x = 1 or 3 gives the statement I asserted. That plus your current third equation evaluated at y=1 gives a much nicer proof. $\endgroup$ – wnoise Apr 17 '18 at 22:16
  • $\begingroup$ @wnoise Can you please expliciteelaborate (write it down) what you are saying. $\endgroup$ – Aqua Apr 18 '18 at 7:57
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    $\begingroup$ @ChristianF: I can't see what wasn't explicit enough about my comment. In order to actually fit more, I've expanded it into an actual answer. $\endgroup$ – wnoise Apr 19 '18 at 0:58

$x$ and $y$ are always explicitly positive, and this will not be further noted.

$$ \begin{align} 2f(xy) &= f(x)f(y)+f(3/x) f(3/y) \tag{Definition} \label{def} \\ f(3) &= 1 \tag{Initial Condition} \end{align} $$

Following @ChristianF, we'll set variables to make the conditions more symmetric.

Considering the case where $y=3/x$, we have: $$ \begin{align} 2f(x)f(3/x) &= 2f(3) = 2 \\ f(3/x) &= \frac{1}{f(x)} \tag{1} \label{1} \\ f(1) &= \frac{1}{f(3)} = 1 \tag{2} \label{2} \end{align} $$

Substituting equation \ref{1} into the original equation $\ref{def}$ (twice: once with $x$ and once with $y$) gives: $$ 2 f(x y) = f(x)f(y)+\frac{1}{f(x) f(y)} \tag{3} \label{3} $$

Considering the case where $y=1$ in equation \ref{3} immediately gives: $$ \begin{align} 2 f(x\cdot 1) &= f(x)f(1)+\frac{1}{f(x)f(1)} \\ 2 f(x) &= f(x)+\frac{1}{f(x)} \\ 2 f(x)^2 &= f(x)^2 + 1 \\ f(x)^2 &= 1 \tag{4} \label{4} \\ \end{align} $$

Then considering the case where $x=y$ in \ref{3}, and simplifying with \ref{4}: $$ \begin{align} 2 f(x^2) &= f(x)f(x)+\frac{1}{f(x) f(x)} \\ 2 f(x^2) &= 1 + \frac{1}{1} \\ f(x^2) &= 1 \\ \end{align} $$

Finally, as $x$ is only positive $x$ and $x^2$ cover the same region: $f(x) = 1$.

  • $\begingroup$ I was courios how you deduce from $f(x)^2=1$ that $f(x)=1$. Thank you. $\endgroup$ – Aqua Apr 19 '18 at 8:54

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