Solution to equation in surd $\sqrt{a+b\sqrt{c}}=\sqrt{x}+\sqrt{y}$
where $a, b, c\in\mathbb{Z}^+$ and x, y $\in \mathbb{Q} $
Please help show how to disprove or prove. Thanks a lot

Sorry about my previous answer, hopefully I've learned to read questions properly.
In the case where $c$ is not a square number there is a criterion.
We have:
$$ \sqrt{a+b\sqrt{c}} = \sqrt{x} + \sqrt{y} \iff a +b\sqrt{c} = x+y+ 2\sqrt{xy} $$
As $c$ is not square then this holds if and only if:
$$ x+y =a \ , \ \sqrt{xy} = \frac{b \sqrt{c}}{2} $$
Now squaring the second equation:
$$ x+y =a \ , \ xy = \frac{b^2 c}{4} $$
Now consider the quadratic equation in $z$ : $(z-x)(z-y) = 0 $.
Expanding this equation gives: $ z^2 - a z + \frac{b^2 c}{4} = 0 $
Applying the quadratic formula gives:
$$ x,y \in \lbrace  \frac{a \pm \sqrt{a^2-b^2c} }{2} \rbrace $$
and therefore in this case a solution exists iff $a^2 - b^2c = n^2 $
