<s>[INST] Upper bound on double series We consider the sum
$$ \sum_{m \in \mathbb Z^2} \frac{1}{(3 m_1^2+3m_2^2+3(m_1+m_1m_2+m_2)+1)^2}. $$
Numerically, it is not particularly hard to see that the value of this series is well below $4$, indeed one gets numerically an upper bound of roughly $3.43$
I wonder if there is analytically a quick argument that the value of this double sum is less than $4$?
EDIT: One could observe that
$$ \sum_{m_1=-1}^1\sum_{m_2=-1}^1 \frac{1}{(3 m_1^2+3m_2^2+3(m_1+m_1m_2+m_2)+1)^2} = \frac{40545}{12544}. $$
So it could suffice to show that the remaining terms are sufficiently small.
 [/INST] As noted in the comment by Beni Bogosel, the sum in question is
\begin{equation}
    s:=\sum_{x=-\infty}^\infty\sum_{y=-\infty}^\infty\frac1{f(x,y)^2},  
\end{equation}
where
\begin{equation}
    f(x,y):=\frac32\, ((x + 1)^2 + (y + 1)^2 + (x + y)^2) - 2. 
\end{equation}
Note that
\begin{equation}
    f(x,y)\ge x^2+y^2+2\ge2\sqrt{x^2+1}\sqrt{y^2+1}\text{ if } \max(|x|,|y|)\ge3. 
\end{equation}
So,
\begin{equation}
    s\le s_9+r_{10},
\end{equation}
where
\begin{equation}
    s_9:=\sum_{x=-9}^9\sum_{y=-9}^9\frac1{f(x,y)^2}<3.42256 
\end{equation}
and
\begin{equation}
\begin{aligned}
    r_k&:=4\sum_{x\ge k}\sum_{y=-\infty}^\infty\frac1{4(x^2+1)(y^2+1)} \\ 
    &=\sum_{x\ge k}\frac1{x^2+1}\sum_{y=-\infty}^\infty\frac1{y^2+1} \\ 
    &\le\int_{k-1/2}^\infty\frac{dx}{x^2}\,\Big(1+2\int_2^\infty\frac{dy}{y^2}\Big)
    =\frac5{k-1/2}<0.52632 
\end{aligned}
\end{equation}
for $k=10$.
Thus,
\begin{equation}
    s<3.42256 + 0.52632<4,
\end{equation}
as desired.
</s>