Prove that $\eta (1/2) \gt \frac{\sqrt 2 \pi^2}{48}$
I am out of any good ideas, but the $\pi^2$ suggests some comparison with $\zeta(2)$.
Here is a bad try:
$$\eta (1/2) \gt 1-\frac{1}{\sqrt 2} + \frac{1}{\sqrt 3}$$
Now it just remains to show that $\frac{\sqrt 2 \pi^2}{48} \lt 1-\frac{1}{\sqrt 2} + \frac{1}{\sqrt 3}$.
Any ideas?
If we are talking about the Dirichlet eta function then try this trick: $$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}=\frac 1{1^s}-\frac1{2^s}+\cdots$$ so: $$\eta(s)-1=-\frac1{2^s}+\frac1{3^s}-\cdots$$ $$2^s\left[\eta(s)-1\right]=-1+\frac{2^s}{3^s}-\cdots=-\eta(s)+2^s(\zeta(s)-1)$$
(Sorry I'm trying to do this in my head) anyway you get the expression: $$\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$$ so if we extend the eta function through an extension of the zeta function we can define its value in terms of $\zeta(1/2)$