Constant functions 
Let $f$ , $g$ , $h$ be three functions from the set of positive real numbers to itself satisfying $$f(x)g(y) = h\left((x^2+y^2)^{\frac{1}{2}}\right)$$ for all positive real numbers $x$ , $y$ . Show that $\dfrac{f(x)}{g(x)}$ , $\dfrac{g(x)}{h(x)}$ and $\dfrac{h(x)}{f(x)}$ are all constant functions .

I have proved that $\dfrac{f(x)}{g(x)}$ is constant and can see that proving either of the last two will prove the final one , but I am not able to prove any of the last two .
Thanks for any help .
 A: forall $x,y,z>0$,  $h(\sqrt{x^2+y^2})f(z) = f(x) g(y) f(z) = f(x) h(\sqrt{y^2+z^2})$,
thus $\frac {h(\sqrt{x^2+y^2})}{h(\sqrt{z^2+y^2})} = \frac{f(x)}{f(z)} $.
Therefore, forall $x,y,z,t > 0$ : $\frac {h(\sqrt{x^2+z^2})}{h(\sqrt{y^2+z^2})} = \frac {f(x)}{f(y)} = \frac {h(\sqrt{x^2+z^2+t^2})}{h(\sqrt{y^2+z^2+t^2})} = \frac {f(\sqrt{x^2+z^2})}{f(\sqrt{y^2+z^2})}$, thus $\frac {h(\sqrt{x^2+z^2})}{f(\sqrt{x^2+z^2})} = \frac{h(\sqrt{y^2+z^2})}{f(\sqrt{y^2+z^2})} $, which proves that $h/f$ is a constant function.
A: We may assume $f(1)=g(1)=1$. It follows that $f(x)=h\bigl(\sqrt{x^2+1}\bigr)=g(x)$ for all $x>0$. Put $$H(t):=h\bigl(\sqrt{t}\bigr)\qquad(t>0)\ ,$$
then
$$f(x)\ f(y)=H(x^2+y^2)\qquad(x>0, \ y>0)\ .$$
Taking logarithms we obtain $\log f(x)+\log f(y)=\log H(x^2+y^2)$
or
$$\log H(x^2+1)+\log H(y^2+1)=\log H(x^2+y^2)\ .\qquad(1)$$
We now write $x^2:=1+u$, $\ y^2:= 1+v$ with $u$ and $v$ near $0$ and introduce the new function $\phi(t):=\log H(2+t)$. Then $(1)$ becomes the familiar functional equation
$$\phi(u)+\phi(v)=\phi(u+v)\ .\qquad(2)$$
If we insist that $f$, $g$, $h$ are continuous then the only solutions to $(2)$  are the functions $\phi(t)=C\,t$, and going all the way backwards the claim about $f$, $g$, $h$ follows.
A: $f , g , h$ are three functions from the set of positive real numbers to itself
Hence $f(0)$ and $g(0)$ are constants.
$$f(x)g(0) = h\left((x^2+0^2)^{\frac{1}{2}}\right)$$
           $$f(x)g(0) = h(x)$$
        $$ \dfrac{h(x)}{f(x)} = g(0)$$
  There fore,    $\dfrac{h(x)}{f(x)}$ is a constant function
$$f(0)g(x) = h\left((0^2+x^2)^{\frac{1}{2}}\right)$$
           $$f(0 )g(x) = h(x)$$
        $$ \dfrac{h(x)}{g(x)} = f(0)$$
     There fore,  $\dfrac{h(x)}{g(x)}$ is a constant function
$$f(x)g(y) = h\left((x^2+y^2)^{\frac{1}{2}}\right)= f(y)g(x)$$
Take $y=0$, $$f(x)g(0) =  f(0)g(x)$$
$$\dfrac{f(x)}{g(x)} = \dfrac{f(0)}{g(0)}$$
 There fore, $\dfrac{f(x)}{g(x)}$ is a constant function
A: We suppose $f,g$, and $h$ are all continuous. I claim that $\lim_{x \rightarrow 0}f(x)$, $\lim_{x \rightarrow 0}g(x)$, and $\lim_{x \rightarrow 0}h(x)$ all exist and are positive. To see this, note that $f(x) = {h(\sqrt{x^2 + 1}) \over g(1)}$ and take limits as $x \rightarrow 0$ on the right. One does the symmetrical argument to show $\lim_{x \rightarrow 0} g(x)$ exists and is positive, and then since $h(x) = f({x \over \sqrt{2}})g({x \over \sqrt{2}})$ we have that $\lim_{x \rightarrow 0} h(x)$ is a positive number as well.
Thus we can extend the domain of definition of all three functions and assume that $f,g,$ and $h$ are continuous positive functions on $[0,\infty)$. By continuity (taking limits as $x$ and/or $y$ go to zero) the functional equation will hold for any $x$ and $y$ in $[0,\infty)$. Now we may plug $y = 0$ into the functional equation and get that $f(x)g(0) = h(x)$, and plugging $x = 0$ into the functional equation we get that $f(0)g(y) = h(y)$ and we are done.
