I'm studying the following function:
$\theta(p)=\eta(p)\eta(p-2)-\frac{p-1}{p}\eta^2(p-1)$,
where $\eta$ - Dirichlet's eta function. If we build a plot for $p\in [1,150]$, it's easy to see that it's positive and decreasing for $p\in [3,150]$. I have a hypothesis that this function remains positive for all $p\ge 1$.
Are there any analytical results on this subject?
You can say stronger things. Since $1-2^{-x}<\eta(x)<1$ for real $x>0$ $$ 1>\eta(p)\eta(p-2)>(1-2^{-p})(1-2^{-p+2})>1-\frac{5}{2^p} \\ 1>\eta(p-1)^2>(1-2^{-p+1})^2>1-\frac{4}{2^p} $$ So $$ -\frac{5}{2^p} < \eta(p)\eta(p-2)-\eta^2(p-1) < \frac{4}{2^p} $$ And for a sequence $\epsilon(p)$ taking on values in $[0,1)$ $$ \epsilon-\frac{4\epsilon+5}{2^p}<\eta(p)\eta(p-2)-(1-\epsilon)\eta^2(p-1) < \epsilon+\frac{4}{2^p} $$ In your case $\epsilon(p)=1/p$, so $$ \frac{1}{p}-\frac{5+4/p}{2^p}<\theta(p)<\frac{1}{p}+\frac{4}{2^p} $$ It follows that $\theta(p)>0$ for $p\ge 5$ and that it converges quickly to $1/p$.