Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction

Question

Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction

I tried to add $\frac{1}{(k+1)^2}$ to both sides of this inequality assuming it's true for $n = k$, but eventually it resulted in a complex expression in the second side:

$$ \frac{3k}{2k+1} + \frac{1}{(k+1)^2} = \frac{3k^3+6k^2+5k+1}{(k+1)^2 (2k+1)} $$

that I'm not sure how to finish the proof using it.

Am I going the right way? If so, how can I finish this proof? and is there an easier way to prove this by induction without facing such confusing expressions?

Thanks.

Answer 1

When $n=2$, it directly works.

Assume it works for $n=k\ge 2$, and try to show that for $n=k+1$.

$1+...+\frac{1}{k^2}>\frac{3k}{2k+1}$ is given, so it suffices to show $\frac{1}{(k+1)^2}\ge\frac{3(k+1)}{2(k+1)+1}-\frac{3k}{2k+1}=\frac{3}{4(k+1)^2-1}\Leftrightarrow (k+1)^2\ge 1$, which is obvious.

Answer 2

Now, $$\frac{3k}{2k+1}+\frac{1}{(k+1)^2}-\frac{3(k+1)}{2k+3}=\frac{k(k+2)}{(k+1)^2(2k+1)(2k+3)}>0$$ and by induction we are done!