A problem regarding the ratios $\frac{f(x)}{g(x)}$ and $\frac{g(x)}{h(x)}$, assuming $f(x)g(y) = h\big(\sqrt{x^2+y^2}\big)$ $\mathbf {The \ Problem \ is}:$ Let, $f,g,h$ be three functions defined from $(0,\infty)$ to $(0,\infty)$ satisfying the given relation $f(x)g(y) = h\big(\sqrt{x^2+y^2}\big)$ for all $x,y \in (0,\infty)$, then show that $\frac{f(x)}{g(x)}$ and $\frac{g(x)}{h(x)}$ are constant.
$\mathbf {My \ approach} :$ Actually, by putting $x$ in place of $y$ and vice-versa, we can show that $\frac{f(x)}{g(x)}$  is a constant, let it be $c .$ Then, I tried that $g(x_i)g(y_i)=g(x_j)g(y_j)$ whenever  $(x_i,y_i)$, $(x_j,y_j)$ satisfies $x^2+y^2 =k^2$ for every $k \in (0,\infty)$ .
But, I can't approach further.
Any help would be greatly appreciated .
 A: Yes you're right at the beginning:
$$P(x,y) \implies f(x)g(y)=h\left(\sqrt{x^2+y^2}\right)$$
$$P(y,x) \implies f(x)g(y)=g(x)f(y)\implies f(x)/g(x)=f(y)/g(y)=c$$
$$\implies f(x)=cg(x) \implies cg(x)g(y)=h\left(\sqrt{x^2+y^2}\right)$$
We can now substitute $y$ with $k$, where $k$ is an infinitesimally small value close to zero from the positive side, getting  $g(x)/h\left(\sqrt{x^2+k^2}\right)=c_2$ for $c_2=1/cg(k)$, and since $x^2 \ge 0$ we have $\sqrt{x^2+k^2} \ge k >0$, if we perceive it as $k=0$ we get $g(x)/h(x) = c_2$ and we're done, otherwise we can believe that $\sqrt{x^2+k^2}$ is infinitesimally close to $x$.
A: To make the formulas look simpler, define the functions $ \tilde f $, $ \tilde g $ and $ \tilde h $ from $ ( 0 , + \infty ) $ to $ ( - \infty , + \infty ) $ by:
$$ \tilde f ( x ) = \log \frac { f \left( \sqrt x \right) } { f ( 1 ) } \qquad \tilde g ( x ) = \log \frac { g \left( \sqrt x \right) } { g ( 1 ) } \qquad \tilde h ( x ) = \log \frac { h \left( \sqrt x \right) } { h \left( \sqrt 2 \right) } $$
Then the functional equation
$$ f ( x ) g ( y ) = h \left( \sqrt { x ^ 2 + y ^ 2 } \right) \tag 0 \label 0 $$
transforms to
$$ \tilde f ( x ) + \tilde g ( y ) = \tilde h ( x + y ) \text , \tag 1 \label 1 $$
together with $ \tilde f ( 1 ) = \tilde g ( 1 ) = \tilde h ( 2 ) = 0 $. Now, substituting $ y $ for $ x $ and $ x $ for $ y $ in \eqref{1} and comparing the result with \eqref{1}, you get $ \tilde f ( x ) + \tilde g ( y ) = \tilde f ( y ) + \tilde g ( x ) $. Letting $ y = 1 $ in the last equation, you have
$$ \tilde f ( x ) = \tilde g ( x ) \text . \tag 2 \label 2 $$Now, use \eqref{1} twice to get
$$ \tilde f ( x + 1 ) + \tilde g ( y ) = \tilde h \big( ( x + 1 ) + y \big) = \tilde h \big( x + ( y + 1 ) \big) = \tilde f ( x ) + \tilde g ( y + 1 ) \text. $$
Letting $ y = 1 $ in the above equation you have $ \tilde f ( x + 1 ) = \tilde f ( x ) + \tilde g ( 2 ) $, which using \eqref{1} and \eqref{2} gives
$$ \tilde g ( x + y ) = \tilde f ( 1 ) + \tilde g ( x + y ) = \tilde h \big( 1 + ( x + y ) \big) = \tilde h \big( ( x + 1 ) + y \big) \\ = \tilde f ( x + 1 ) + \tilde g ( y ) = \tilde f ( x ) + \tilde g ( 2 ) + \tilde g ( y ) = \tilde g ( x ) + \tilde g ( y ) + \tilde g ( 2 ) \text . $$
Substituting $ \frac x 2 $ for both $ x $ and $ y $ in the above equation you get $ \tilde g ( x ) = 2 \tilde g \left( \frac x 2 \right) + \tilde g ( 2 ) $. Now, using \eqref{1} and \eqref{2} you get
$$ \tilde h ( x ) = \tilde h \left( \frac x 2 + \frac x 2 \right) = \tilde f \left( \frac x 2 \right) + \tilde g \left( \frac x 2 \right) = 2 \tilde g \left( \frac x 2 \right) \text . $$
This, together with the previous result shows that
$$ \tilde g ( x ) = \tilde h ( x ) + \tilde g ( 2 ) \text . \tag 3 \label 3 $$
You can now rewrite \eqref{2} and \eqref{3} in terms of the original functions and get
$$ \frac { f ( x ) } { g ( x ) } = \frac { f ( 1 ) } { g ( 1 ) } $$
and
$$ \frac { g ( x ) } { h ( x ) } = \frac { g \left( \sqrt 2 \right) } { h \left( \sqrt 2 \right) } \text , $$
as desired.
Of course, you could avoid defining $ \tilde f $, $ \tilde g $ and $ \tilde h $, and use \eqref{0} and some messier equations corresponding to the above ones, in terms of $ f $, $ g $ and $ h $. But I find this way more elegant, and in fact I think this way you can see the simple idea behind the solution more easily.
