How to find x such that $\sqrt{x}= \sqrt{a-x} + \sqrt{b-x} + \sqrt{c-x}$? Let $a,b,c \in R$ are given. How to find x such that $\sqrt{x}= \sqrt{a-x} + \sqrt{b-x} + \sqrt{c-x}$? Is there a simple way?

As I wrote in a comment, to get a polynomial in $x$, you must square multiple times. For starting $$(\sqrt{x}- \sqrt{a-x})^2=(\sqrt{b-x} + \sqrt{c-x})^2$$ $$a-2 \sqrt{x} \sqrt{a-x}=2 \sqrt{b-x} \sqrt{c-x}+b+c-2 x$$ $$a-b-c+2 x=2 \sqrt{x} \sqrt{a-x}+2 \sqrt{b-x} \sqrt{c-x}$$ Squaring again $$(a-b-c+2 x)^2=4(\sqrt{x} \sqrt{a-x}+ \sqrt{b-x} \sqrt{c-x})^2$$ Continue the same way, pushing everytime the radicals to the lhs. 
Hoping no mistakes (not sure !), you should end with something like $$80 x^4-64 (a+b+c)x^3 +8  \left(a^2+6 a (b+c)+b^2+14 b c+c^2\right)x^2-64 a b c x+\left(a^2-2 a (b+c)+b^2+6 b
   c+c^2\right)^2=0$$ Now, you have a quartic polynomial ... which can be solced using radicals.
I wish you a very good time !!
By the way, do not forget that squaring introduces extra roots and some ot them need to be discarded later after checks (in your case, if solution exists it must be between $0$ and the minimum of $a,b,c$).
