To prove given function is constant function Suppose $f,g,h$ are functions from the set of positive real numbers into itself satisfying
$$f(x)g(y)=h\left(\sqrt {x^2+y^2} \right)\ \ \forall \ x,y\in (0,\infty )$$
Show that the three functions $\frac {f(x)}{g(x)},\frac{g(x)}{h(x)},\frac{h(x)}{f(x)}$ are all constants.
I only succeed in to show $f/g$ is constant function i,e, i got $f(x)=g(x)$.
Anyone can help me to prove that $g/h$ is constant.
 A: So, by dividing by $f(y)g(x)$, you get $f(x)/g(x)=f(y)/g(y)$ for all positive $x,y$. So then obviously $f/g$ is constant.
Ok, here is my try: $f(x)g(1)=h(\sqrt{x^2+1})$ and $f(x)g(\sqrt{2})=h(\sqrt{x^2+2})=f(\sqrt{x^2+1})g(1)$. 
So,$f(x)g(\sqrt{2})=f(\sqrt{x^2+1})g(1)$. i.e. $f(x)=f(\sqrt{x^2+1})g(1)/g(\sqrt{2})$. Hence $f(\sqrt{x^2+1})/h(\sqrt{x^2+1})=g(\sqrt{2})/g(1)^2$. Now change variable $y=\sqrt{x^2+1}$. But we are done only for $y>1$.
A: Assume that you have two positive numbers $u$ and $v$ and select $c$ smaller than both. Then you have that you can write $u=\sqrt{x^2+c^2}$ and $v=\sqrt{y^2+c^2}$ for some positive $x$ and $y$
$$f(x)g(c) = h\left(\sqrt{x^2+c^2}\right)$$
$$f(x)g(c\sqrt2) = h\left(\sqrt{x^2+2c^2}\right) =
   f\left(\sqrt{x^2+c^2}\right)g(c)$$
So 
$${h(u)\over f(u)} = {h\left(\sqrt{x^2+c^2}\right)\over f\left(\sqrt{x^2+c^2}\right)} = {f(x)g(c)\over f(x)g(c\sqrt2)/g(c)} = g(c)^2/g(\sqrt c)$$
In similar way you get
$${h(v)\over f(v)} = {g(c)^2 \over \sqrt c} = {h(u)\over f(v)}$$
That is $h(v)/f(v) = h(u)/f(u)$ for any positive $u$ and $v$ which means that $h/f$ is constant.
