Inequalities involving geometry but I can't post a picture yet How do I show that
$$ \frac 12 \left(\frac 1 {3^2}+\frac 1{4^2}+ \frac 1{5^2}+\dots\right) < \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots \quad ?$$

After moving the odd terms from the LHS to the RHS, we obtain the following equivalent inequality,
$$\frac 12 \left(\frac 1{4^2}+ \frac 1{6^2}+ \frac 1{8^2}+\dots\right) < \left(1-\frac 12\right)\left( \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots\right).$$
Then note that for all positive integer $n$, each term $\dfrac{1}{(2n)^2}$ is less than $\dfrac{1}{(2n-1)^2}$
