Is this sort of "infinite ascent" a valid proof that $a + b \le 1 + ab$? Today I had to prove that for $a,b \in [0,1]$, we have 
$$a + b \le 1 + ab$$
The first proof I found feels rather weird (as in, there should be a much simpler process) and I am not entirely sure that it's correct. Anyhow, I thought:I can square both sides and see that the inequality is true if and only if
$$a^2 + b^2  \le 1 + a^2b^2 $$
But this is in the same form as the previous inequality (setting $a_1 = a^2$ and $b_1 = b^2$). I can iterate this process and prove that the original inequality is equivalent to 
$$a^n + b^n \le 1 + a^nb^n$$for every $n = 2^k, k \in \mathbb N$ . Now eventually I will find an $n$ such that $a^n, b^n, a^nb^n$ are all small enough such that the inequality holds, hence the first inequality holds, too, and we proved 
$$a + b \le 1 + ab$$
Is this logically sound? (And what is a more immediate proof of this)
 A: The most immediate proof is 
$$a+b \leq 1+ab \Longleftrightarrow (1-a)(1-b) \geq 0$$ which is obvious.
And in general,
$$a^n+b^n \leq 1+a^nb^n \Longleftrightarrow (1-a^n)(1-b^n) \geq 0$$
A: Assume $a,b < 1$ (otherwise the inequality is obvious). Then there is $n$ with $a^n, b^n \leq \frac{1}{2}$. Take $k$ with $2^k \geq n$. Now we have
$$a^{2^k} + b^{2^k} \leq a^n + b^n \leq \frac{1}{2} + \frac{1}{2} = 1 \leq 1 + a^{2^k} b^{2^k}.$$
The equivalence you've shown yields the result. So yes, your idea should work.
A: Yes, your logic is sound. This is because given any $a,b \in [0,1]$ you can find an 'if and only if' argument with a finite number of steps. 
Note that the right hand side of the inequality you're proving is greater than $1$, hence it suffices to find $k$ such that $a^n + b^n <1$ where $n = 2^k$. Once you've done this you can use your 'iff'/infinite ascent argument  $k$ times to conclude that the inequality holds for those specific $a$ and $b$. As this holds for any pair $a,b \in [0,1]$ you're done.
Hope this helps. I like this, it's an interesting way of proving this inequality! 
