Inequalities involving geometry but I can't post a picture yet How do I show that
$$ \frac 12 \left(\frac 1 {3^2}+\frac 1{4^2}+ \frac 1{5^2}+\dots\right) < \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots \quad ?$$
 A: 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 $\dfrac{1}{(2n-1)^2}$
A: Alternatively: Note that the RHS:
$$\frac{\pi ^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots =\\
\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots \right)+\frac{1}{2^2}\left(\underbrace{1+\frac{1}{2^2}+\frac{1}{3^2}\cdots}_{\frac{\pi^2}{6}} \right) \Rightarrow $$
$$\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots=\frac{\pi^2}{6}-\frac{1}{2^2}\cdot\frac{\pi^2}{6}-1=\frac{\pi^2}{8}-1.$$
The LHS:
$$\frac 12 \left(\frac 1 {3^2}+\frac 1{4^2}+ \frac 1{5^2}+\dots\right)=\frac12\cdot\left(\frac{\pi^2}{6}-1-\frac{1}{2^2}\right)=\frac{\pi^2}{12}-\frac{5}{8}.$$
Hence:
$$\frac{\pi^2}{12}-\frac{5}{8}<\frac{\pi^2}{8}-1 \iff 9<\pi^2.$$
