Since you wanted the proof by induction:
Note we adapt the usual procedure of induction: Instead of assuming the proposition is true for an arbitrary value of n implies it is true for (n+1), we show if the proposition is true for BOTH n AND (n+1) then it is true for (n+2). Of course the Base Case now will require us to verify the proposition is true for two consecutive values of n.
Base Case: Clearly 3=1+2 so the claim L(n)=F(n-1)+F(n+1) is certainly true when n=2. I leave you to show that the claim is also true when n=3 (this is very easy).
Assume for some natural number 'n' that
Add the two equations:
L(n)+L(n+1)=[F(n)+F(n-1)] + [F(n+1)+F(n+2)]
but by definition L(n)+L(n+1) = L(n+2)
and similarly F(n)+F(n-1) = F(n+1)
So we deduce L(n+2)=F(n+1)+F(n+3), hence the inductive step is complete.
I hope you understand the proof by induction?