In one-step induction, why sometimes there are 2 base cases? I noticed in various induction proofs, that it uses 2 base cases (e. g: for $n = 1$ and $2$ the problem trivially holds...)
Why is that? Isn't one base case enough for one-step induction?

For example.

Prove that:
  $$\frac{1}{n+1}+\frac{1}{n+3}+\frac{1}{n+5}+...+\frac{1}{3n-1}\geq\frac{1}{2}$$ for any natural $n$.

For the proof by induction we need to check $n=1$ and $n=2$ because we need a transition from $n$ to $n+2$.
I'll write a full proof and I hope it would be clear. 
For $n=1$ we need to check:
$$\frac{1}{2}\geq\frac{1}{2},$$ which is true.
For $n=2$ we need to check: 
$$\frac{1}{3}+\frac{1}{5}\geq\frac{1}{2}$$ or
$$\frac{8}{15}\geq\frac{1}{2},$$ which is true.
Now, it's enough to prove that: $$\frac{1}{n+1}+\frac{1}{n+3}+\frac{1}{n+5}+...+\frac{1}{3n-1}\geq\frac{1}{2}\Rightarrow$$
$$\Rightarrow\frac{1}{n+3}+\frac{1}{n+5}+...+\frac{1}{3n-1}+\frac{1}{3n+1}+\frac{1}{3n+3}+\frac{1}{3n+5}\geq\frac{1}{2},$$ for which it's enough to proof that:
$$\frac{1}{2}-\frac{1}{n+1}+\frac{1}{3n+1}+\frac{1}{3n+3}+\frac{1}{3n+5}\geq\frac{1}{2}$$ or
$$\frac{1}{3n+1}+\frac{1}{3n+5}\geq\frac{2}{3(n+1)},$$ which is true by C-S:
$$\frac{1}{3n+1}+\frac{1}{3n+5}\geq\frac{(1+1)^2}{3n+1+3n+5}=\frac{2}{3(n+1)}$$ and by induction we are done! 
