Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$
For your induction step, all you need is $$ 2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1}. $$ That's equivalent to $$ \frac{1}{k}-\frac{1}{(k+1)^2}-\frac{1}{k+1}\geq 0 $$ i.e. $$ \frac{(k+1)^2-k-k(k+1)}{k(k+1)^2}=\frac{1}{k(k+1)^2}\geq 0. $$ So it holds.
$\displaystyle \frac{1}{i^2}< \frac{1}{i(i-1)}=\frac{1}{i-1}-\frac{1}{i}$,for $i\ge 2$
So we have $\displaystyle 1+\sum_{i=2}^{k}\frac{1}{i^2}\le 1+\sum_{i=2}^{k}(\frac{1}{i-1}-\frac{1}{i})=2-\frac{1}{k}$
I think this is better than induction.