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In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is $\theta$. Then the area of the triangles will be $\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta$. This will be true on both sides. Therefore we can maximise this area by setting $\theta = \frac{\pi}{2}$.

\begin{center}
    \begin{tikzpicture}[scale=2]
        \coordinate (A) at (-1,0.5);
        \coordinate (B) at (1,-0.5);

\coordinate (C) at (-1,-0.5);
        \coordinate (D) at (1,0.5);

\def\d{0.25};

\draw (A) -- (B);
        \draw (C) -- (D);

\def\ka{1/sqrt(3)};
        \def\kb{sqrt(2)/sqrt(3)};

\draw[dashed] ($(A)+{\d}*({\ka},{\kb})$) --  ($(B)+{\d}*({\ka},{\kb})$);
        \draw[dashed] ($(A)-{\d}*({\ka},{\kb})$) --  ($(B)-{\d}*({\ka},{\kb})$);

\filldraw[color=blue, opacity=0.2] ($(A)+{\d}*({\ka},{\kb})$) --  ($(B)+{\d}*({\ka},{\kb})$) --  ($(B)-{\d}*({\ka},{\kb})$) --  ($(A)-{\d}*({\ka},{\kb})$);

\def\kc{1/sqrt(3)};
        \def\kd{-sqrt(2)/sqrt(3)};

\draw[dashed] ($(C)+{\d}*({\kc},{\kd})$) --  ($(D)+{\d}*({\kc},{\kd})$);
        \draw[dashed] ($(C)-{\d}*({\kc},{\kd})$) --  ($(D)-{\d}*({\kc},{\kd})$);

\filldraw[color=blue, opacity=0.2]  ($(C)+{\d}*({\kc},{\kd})$) --  ($(D)+{\d}*({\kc},{\kd})$) -- ($(D)-{\d}*({\kc},{\kd})$) -- ($(C)-{\d}*({\kc},{\kd})$);
    \end{tikzpicture}
\end{center}

Consider the (darker) shaded area. This is our set $A$.

The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider $c_1 = 0, c_2 = 0$, so our lines meet at the origin.

Now also consider the linear transformation $\begin{pmatrix} a_1  & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}$ which takes the coordinate axes to these lines. This will take the square $[-\delta, \delta] \times [-\delta, \delta]$ which has area $4\delta^2$ to our square, which will have area $\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}$.

If we consider the length of the two diagonals of this area, $l_1, l_2$ we know that $\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}$, if we consider the larger of $l_1$ and $l_2$ (wlog $l_1$) we must have that $\frac{l_1^2}{2} \geq  \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}$ and so points on opposite ends of the diagonal will satisfy the inequality in the question.

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