TITLE: Has this equation appeared before?
QUESTION [6 upvotes]: I want to know if the following equation has appeared in mathematical literature before, or if it has any important significance.
$$\sqrt{\frac{a+b+x}{c}}+\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}},$$
where $a,b,c$ are any three fixed positive real and $x$ is the unknown variable.

REPLY [2 votes]: This provides the explicit polynomial in $x$ (for those curious), though I'm not aware if the equation has appeared in the mathematical literature. We get rid of the square roots by multiplying out the $8$ sign changes,
$$\prod^8 \left(\sqrt{\frac{a+b+x}{c}}\pm\sqrt{\frac{b+c+x}{a}}\pm\sqrt{\frac{a+c+x}{b}}\pm\sqrt{\frac{a+b+c}{x}}\right)=0$$
then collecting powers of $x$. It turns out the $8$th-deg equation factors into a linear (cubed), a quadratic, and a cubic. For simplicity, let,
$$\begin{aligned}
p &= a+b+c\\
q &= ab+ac+bc\\
r &= abc
\end{aligned}$$
Then,
$$(p+x)^3=0\tag1$$
$$r^2 - 2 q r x + (q^2 - 4 p r) x^2 = 0\tag2$$
$$p r^2 + r (-2 p q + 9 r) x + (p q^2 - 4 p^2 r + 6 q r) x^2 + (q^2 - 4 p r) x^3 = 0\tag3$$

Example: 

Let $a,b,c = 1,2,4$, then
$$(7+x)^3=0\\
-16 + 56 x + 7 x^2 = 0\\ 
-112 + 248 x - 119 x^2 + 7 x^3 = 0$$
The roots of the quadratic solve,
$$\sqrt{\frac{a+b+x}{c}}\pm\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{a+c+x}{b}}-\sqrt{\frac{a+b+c}{x}}=0$$
while a root of the cubic solves,
$$\sqrt{\frac{a+b+x}{c}}-\sqrt{\frac{b+c+x}{a}}-\sqrt{\frac{a+c+x}{b}}+\sqrt{\frac{a+b+c}{x}}=0$$
and two others, while the linear root takes care of the remaining three sign changes.