I chose to use mathematical induction. We are given to prove:
$$\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le\frac{5}{3}-\frac{2}{2n+1}$$
1.) Confirm the base case $P_1$:
$$\sum_{k=1}^1\left(\frac{1}{k^2}\right)\le\frac{5}{3}-\frac{2}{2(1)+1}$$
$$1\le\frac{5}{3}-\frac{2}{3}=1$$
The base case is true
2.) State the induction hypothesis $P_n$:
$$\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le\frac{5}{3}-\frac{2}{2n+1}$$
3.) Formulate the inductive step:
As our inductive step, we see we need to add to the left side:
$$\frac{1}{(n+1)^2}$$
And to the right side, we need to add:
$$\left(\frac{5}{3}-\frac{2}{2(n+1)+1}\right)-\left(\frac{5}{3}-\frac{2}{2n+1}\right)$$
Simplifying this, we obtain:
$$\frac{5}{3}-\frac{2}{2(n+1)+1}-\frac{5}{3}+\frac{2}{2n+1}$$
$$\frac{2}{2n+1}-\frac{2}{2n+3}=2\left(\frac{(2n+3)-(2n+1)}{(2n+1)(2n+3)}\right)=\frac{4}{(2n+1)(2n+3)}$$
So, we need to be able to show:
$$\frac{1}{(n+1)^2}\le\frac{4}{(2n+1)(2n+3)}$$
Because all factors in the denominator are positive, we may multiply through to obtain:
$$(2n+1)(2n+3)\le4(n+1)^2$$
Expand:
$$4n^2+8n+3\le4n^2+8n+4$$
or:
$$3\le4$$
This is true, and so we may state:
$$\frac{1}{(n+1)^2}\le\frac{4}{(2n+1)(2n+3)}$$
or equivalently:
$$\frac{1}{(n+1)^2}\le\left(\frac{5}{3}-\frac{2}{2(n+1)+1}\right)-\left(\frac{5}{3}-\frac{2}{2n+1}\right)$$
Now, adding this to $P_n$ we get:
$$\sum_{k=1}^n\left(\frac{1}{k^2}\right)+\frac{1}{(n+1)^2}\le\frac{5}{3}-\frac{2}{2n+1}+\left(\frac{5}{3}-\frac{2}{2(n+1)+1}\right)-\left(\frac{5}{3}-\frac{2}{2n+1}\right)$$
Which we may simplify as follows:
$$\sum_{k=1}^{n+1}\left(\frac{1}{k^2}\right)\le \frac{5}{3}-\frac{2}{2(n+1)+1}$$
We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction.
