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.
 A: 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 $m=2,n=7$
