<s>[INST] Following Fibonacci Prove that $ F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}} $?
I discovered this property from an attempt to solve the following problem:

Defining $ A_n = \sqrt{{F_n} ^ 2 + {F_{n + 2}} ^ 2}, $ the numbers $ A_n, $ $ A_ {n + 1} $ and $ A_{n + 2} $ are the length measures of the sides of a triangle whose area is $ \frac{1}{2} $ unit.
  (I also put this in the Wikipedia article about the Fibonacci Sequence).

 [/INST] To prove 
$F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}}
$,
I would write it as
$F^2_{n + 2}(F_n ^ 2 + F_ {n + 1}^ 2)
 = F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + 1 
$
or
$(F_{n+1}+F_n)^2(F_n ^ 2 + F_ {n + 1}^ 2)
 = F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + 1 
$.
For $n=0$, 
$F_n = 0, F_1 = 1$,
this becomes
$(1)^2(1)
 = 1 
$,
which is true.
For $n=1$, 
$F_n = 1, F_{n+1} = 1$,
this becomes
$(2)^2(2)
 = (3+4) + 1 
$,
which is true.
For $n=2$, 
$F_n = 1, F_{n+1} = 2$,
this becomes
$(2+1)^2(1 ^ 2 + 2^ 2)
 = 2 ^ 2(3  +4\cdot 2) + 1 
$,
or
$45
=45
$
which is true.
For $n=3$, 
$F_n = 2, F_{n+1} = 3$,
this becomes
$(3+2)^2(2 ^ 2 + 3^ 2)
 = 2\ 3 ^ 2(3\ 2+4\ 3) + 1 
$.
or
$25\ 13
=18(18)+1
$
or
$325 = 325
which is true.
So, it looks good.
The next step would be
to expand the sides,
eliminate any common terms,
and try to prove it by induction.
I'll leave it at this.
</s>