There is a specific part of the section of Serge Lang's Algebra from Chapter IV Section 7 regarding an inequality preceding the generalized Szpiro Conjecture that confused me. How exactly are the first two inequalities derived from the abc conjecture, and how does the third one follow? Any help would be appreciated, thanks.
The section in question.
The specific form of the abc conjecture used if needed.
Well, the third inequality follows by applying the triangle inequality to $Au^m + Bv^n = k$ and using the previous two.
To get the first two is a bit more trouble - actually, I have something slightly different, but almost the same, so there could be a mistake. First, we may assume $Au^m$ and $Bv^n$ and $k$ are relatively prime; in principle this is not true if $A$ and $B$ have a common divisor, but we may factor it out and ignore it at the cost of worsening the constant in the $<<$, since it only depends on the fixed values $A$ and $B$. Applying $abc$ directly, $$\max \{ |Au^m|, |Bv^n|, |k|\} << N_0 (Au^m Bv^n k)^{1+\epsilon}$$ Observe that changing the powers on $u$ and $v$ does not change the radical (I suppose I am taking them to be positive here), and we can drop $A$ and $B$ at the cost of increasing the constant. That leaves us with $N_0(uvk)$ inside, and the radical is multiplicative for coprime numbers, and $N(x)< |x|$, so $$\max \{ |Au^m|, |Bv^n|, |k|\} << |u||v|N_0(k)^{1+\epsilon}$$ The LHS is obviously larger than either $|Au^m|$ or $|Bv^n|$ and we can again drop $A$ and $B$. This leaves $$|u|^m << |u||v|N_0(k)^{1+\epsilon}$$ $$|v|^n << |u||v|N_0(k)^{1+\epsilon}$$
In the first, isolate $u$, and in the second, $v$: $$|u|^{m-1} << |v|N_0(k)^{1+\epsilon}$$ $$|v|^{n-1} << |u|N_0(k)^{1+\epsilon}$$
Take an $n-1$st power of the first, then substitute the second $$|u|^{(m-1)(n-1)} << |u|N_0(k)^{(n-1)(1+\epsilon)}$$
Moving the straggler $|u|$ over, we get $$|u|^{mn -m - n} << N_0(k)^{(n-1)(1+\epsilon)}$$ and hence $$|u| << N_0(k)^{\frac{n-1}{mn - m - n}(1+\epsilon)}$$
This is identical to what Lang writes, except I have $n-1$ in the numerator of the exponent rather than $1$.
A symmetric argument works for the other inequality.