wtf

everything follows logically using the identity

(a)+(1/a) >= 2

For [math] n=2 [/math] you have
[math] a_2 = 2^y [/math]

and [math] g(n) = y^1 = y [/math] so
[math] 2^(g(2)) = 2^y [/math]

So it holds for [math] n=2 [/math] Now suppose it is true up to [math] k [/math]. That means [math] a_k >= 2^{(y^{(k-1)})} [/math] and then consider [math] n=k+1 [/math]

[math] a_{k+1} = (a_k + 1/(a_k))^y [/math]

and [math] g(k+1) = y^k [/math]
so [math] 2^(g(k+1)) = 2^{(y^k)} [/math]

now suppose that instead [math] a_{k+1} < 2^{(y^k)} [/math]

Take the yth root of both sides to get
[math] a_k + 1/(a_k) < 2^{(y^{(k-1)})} [/math]

But that is a contradiction, because we know that [math] a_k [/math] alone is larger than [math] 2^{(y^{(k-1)})} [/math], given our inductive hypothesis. So instead the larger or equal than relation must be true, by contradiction.

Therefore this is true for all [math] n>= 2 [/math], by induction.

>>8508444
What you're trying to say in the most convoluted way possible is that
[eqn]a_{k+1}=(a_k+1/a_k)^\lambda\geq a_k^\lambda\geq (2^{\lambda^{k-1}})^\lambda=2^{\lambda^{k-1}\lambda}=2^{\lambda^k}[/eqn]

>>8508467
Yeah, but those equalities and inequalities are not immediately provable. That is why I turned to a contradiction argument.

Could it be proven directly? Provably. Do I want to waste my time to find the exact train of logic needed for that? Not at all. The proof by contradiction of the inductive step is so easy and obvious.

>>8508470
Surely the implications of taking the y'th root are less obvious than anything I did?

>>8508467
Reading your entire thing closely again I see that it is all justified so your direct proof is also correct.

Though to me the contradiction was more obvious than doing this. It is the first thing that came to my mind.

>>8508483
Contradiction is a great tool but I guess using it unnecessarily is a pet peeve of mine, although I feel like it's a sentiment shared by more than me.

>>8508437
> a=0
LOLL!!!!