Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$ 
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$

I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know how simplify it to 2.
I try:
Let $a=F_n$, $b=F_{n+1}$ and $c=F_{n+2}$
$a^4+b^4+c^4+2(ab)^2+2(ac)^2+2(bc)^2=2a^4+2b^4+2c^4$
$2(ab)^2+2(ac)^2+2(bc)^2=a^4+b^4+c^4$
I am not sure, what to do next. Can anyone help by completing the prove?

As $F_{n+2}=F_{n+1}+F_n$. One first solves the characteristic polynomial. 
$$
X^2=X+1
$$
on gets two roots $r=\frac{1+\sqrt{5}}{2}$ and $\bar{r}=\frac{1-\sqrt{5}}{2}$. 
One now solves $\alpha.r^n+\beta.\bar{r}^n=F_n$ for the two first indices.
The OP did not indicate his initialization so I suppose that $F_0=0;\ F_1=1$, one gets then $F_n=\frac{1}{\sqrt{5}}(r^n-\bar{r}^n)$. This, with the relations $r+\bar{r}=1;\ r.\bar{r}=-1;\ r-\bar{r}=\sqrt{5}$ will, I think, be sufficient to get the desired result. 
Coda In fact, it is true in general that 
$$
\Big(X^2+Y^2+(X+Y)^2\Big)^2=2\Big(X^4+Y^4+(X+Y)^4\Big)
$$
thus, the fraction holds true for any sequence s.t. $s_{n+2}=s_n+s_{n+1}$.   
