How do you find the value of $m$ and $n$ if $x+y+z=\frac{m}{\sqrt n}$ given certain conditions on x,y,z? Problem:
Let $x,y$ and $z$ be real numbers satisfying:
$$x=\sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}}$$
$$y=\sqrt{z^2 - \frac{1}{25}} + \sqrt{x^2 - \frac{1}{25}}$$
$$z=\sqrt{x^2 - \frac{1}{36}} + \sqrt{y^2 - \frac{1}{36}}$$
If $x+y+z=\frac{m}{\sqrt n}$, where $m,n \in N$ and $n$ is not divisible by the square of any prime number, then find $m$ and $n$

I don't know how to even begin with this question, so any help will be appreciated.

Working on The hint of joey we can make this triangle and take x,y,z as the sides of triangle as follows without the loss of generality....

Now just to make a detailed figure ....

Just to sum these up ...from the figure we have ...
$$DC=x=\sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}}$$
$$BD=y=\sqrt{z^2 - \frac{1}{25}} + \sqrt{x^2 - \frac{1}{25}}$$
$$BC=z=\sqrt{x^2 - \frac{1}{36}} + \sqrt{y^2 - \frac{1}{36}}$$

Now note an interesting thing...the triangles $ABC,ACD,ABD$ all have equal sides that is they are congruent so there areas must also be equal ... So 
Area of ABC $\frac{1}{2}×z×\frac{1}{6}$ equal to($=$) 
area of ACD $\frac{1}{2}×x×\frac{1}{4}$ equal to($=$)
area of ABD $\frac{1}{2}×y×\frac{1}{5}$ equal to ($=$)
That is we can say ... $$\frac{1}{2}×y×\frac{1}{5} = \frac{1}{2}×z×\frac{1}{6}=\frac{1}{2}×x×\frac{1}{4}$$ solving which gives $x+y+z=\frac{15x}{4}$ ...now we just need to find $x$ ...any ideas from your side...
Edit: Thanks to the hint of Joey that is also easily done by using herons formula and then equating it to the (area of ACD $=1/2×x×1/4$ ) 

After calculating the area by Herons formula as shown in the image you will get Area as $\frac{15x^2\sqrt{7}}{64}$ which is equal to $1/2×x×1/4$ solving this you will get $x=8/15\sqrt{7}$ put this in the value of $x+y+z$ to get answer $2/\sqrt{7}$ so answer is $m=2,n=7$
