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Let the following four distinct lines be given, $$ \begin{aligned} L_0&:\; 113x - 994y + 24 = 0,\\ L_1&:\; 459x - 888y + 967 = 0,\\ L_2&:\; -828x - 561y - 620 = 0,\\ L_3&:\; -182x - 983y - 1769 = 0. \end{aligned} $$

Each triple of these lines bounds an obtuse triangle.
Let $Q=L_0L_1L_2L_3$ be the quadrilateral of the four lines. enter image description here

Polar circles and the Newton line

For any three lines $L_i,L_j,L_k$ forming a triangle with vertices $A,B,C$:

  • Let $H$ be the orthocenter,
    $R$ the circumradius,
    and $r^2 = 4R^2 - \tfrac12(a^2+b^2+c^2)$ its polar-circle radius squared.

Then the polar circle of the triple $(L_i,L_j,L_k)$ is the circle centered at $H$ of radius $\sqrt{r^2}$.

For the four lines $L_0,\dots,L_3$, the two diagonals of $Q$ meet at points whose midpoints $M_1,M_2$ determine the Newton line of the quadrilateral.

Observation

For the above quadruple:

  • all four triangles $(L_i,L_j,L_k)$ are obtuse, so all four polar circles are real (positive $r^2$);
  • for the triple $(L_0,L_1,L_2)$, the Newton line does not meet its polar circle in any real point.

From https://math.stackexchange.com/a/5100505, no member of the inscribed-conic pencil has eccentricity $\sqrt2$. Can you prove that eccentricities of all hyperbolic members lie below $\sqrt{2}$?

For a general quadrilateral of lines $Q$ in which every triple forms an obtuse triangle, the family of conics tangent to the four sides has centers along the Newton line.

Question

How can one prove the eccentricity among the real members of this family attains maximum $e_{\max}(Q)$ and satisfies

$$ e_{\max}(Q) < \sqrt{2} $$

if every triple of sides of $Q$ is obtuse and Newton line does not meet its polar circle in any real point?

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I studied your example and can confirm that the eccentricities of all the inscribed conics is less than $\sqrt2$.

Define: $A=L_0\cap L_1$, $B=L_0\cap L_2$ and $E=A+t(B-A)$. The pencil of inscribed conics can be parametrised by choosing $E$ as the tangency point on line $L_0$. One can then find the equation of a generic conic in the pencil as a function of $t$, and write an expression for its eccentricity $e(t)$.

I made the computation with Mathematica and found: $$ e^2(t)=2\left( 1+{d(t)\over\sqrt{r(t)}} \right)^{-1}, $$ where: $$ d(t)=d_0 + d_1 t + d_2 t^2, \quad r(t)=r_0 - r_1 t + r_2 t^2 + r_3 t^3 + r_4 t^4, $$ and $d_i$, $r_i$ are (large) positive integers. Both $d(t)$ and $r(t)$ don't have real roots and are always positive, so that $e^2(t)$ is a smooth function, with a maximum $\approx 1.17259$ (at $t\approx -0.190334$) and $$ \lim_{t\to+\infty}e^2(t)=\lim_{t\to-\infty}e^2(t)\approx1.03206. $$ Here's a plot of $e^2(t)$ in the range $(-5,5)$:

enter image description here

And here's an animated diagram, made with GeoGebra:

enter image description here

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