Steiner inellipse Hello it's related to my answer for Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq 3\frac{a+b+c}{ax+by+cz}$ 
My answer fails but I don't know why ... So I was thinking a generalization of the following formula:
$$\frac{IA^2}{CA\cdot AB}+\frac{IB^2}{BC\cdot AB}+\frac{IC^2}{CA\cdot BC}=1$$
I know that it's related to the Steiner inellipse and we have for a triangle ABC and the ellipse of foci $P$ and $Q$:
$$\frac{PA\cdot QA}{BA\cdot CA}+
  \frac{PB\cdot QB}{CB\cdot AB}+
  \frac{PC\cdot QC}{BC\cdot AC}=1$$
But in my proof I have also use the following formula:
\begin{align}
  \frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2} &=
  \frac{1}{r^2}-\frac{1}{2rR} \\
  IA^2+IB^2+IC^2 &= s^2+r^2+8rR \\
  CA\cdot AB+BC\cdot AB+CA\cdot BC &= s^2+(4R+r)r \\
  \frac{1}{CA\cdot AB}+\frac{1}{BC\cdot AB}+\frac{1}{CA\cdot BC} &=
  \frac{1}{2rR}
\end{align}
So what's the new expression of:
\begin{align}
  \frac{1}{BA\cdot CA}+\frac{1}{CB\cdot AB}+\frac{1}{BC\cdot AC} &= ? \\
  \frac{1}{PA\cdot QA}+\frac{1}{PB\cdot QB}+\frac{1}{PC\cdot QC} &= ?  \\
  PA\cdot QA+PB\cdot QB+PC\cdot QC &= ?  \\
  BA\cdot CA+CB\cdot AB+BC\cdot AC &= ?
\end{align}
In function of the parameters of the inellipse and the triangle $ABC$ like the area and the side of the triangle or the semi major semi minor axes of the ellipse?
Edit: I have a good news
The centroid $M$ of the triangle $ABC$ correspond to the centre of the inellipse  and we have the following relation for $P$ any interior point  related to the triangle $ABC$:
$$PA^2+PB^2+PC^2=MA^2+MB^2+MC^2+3MP^2$$

Thanks a lot.
 A: A triangle $\triangle ABC$ with its centroid at the origin is the image of an origin-centered equilateral under a linear transformation. The transformation carries the equilateral's incircle to $\triangle ABC$'s Steiner inellipse. With an appropriate rotation, and by considering similar triangles (and/or ellipses) equivalent, we may assume that the transformation is a simple vertical scaling. This being so, ...

Consider the origin-centered equilateral $\triangle A_\star B_\star C_\star$ with vertex coordinates
$$A_\star := 2 r \operatorname{cis}\theta \qquad B_\star := 2 r\operatorname{cis}(\theta+120^\circ) \qquad C_\star := 2 r \operatorname{cis}(\theta-120^\circ)$$
where $\operatorname{cis}(\cdot) := (\cos(\cdot), \sin(\cdot))$. The (origin-centered) incircle of this triangle has radius $r$. For some non-negative $s<r$, apply the vertical-scaling transformation
$$(x,y) \to \left(\;x, \frac{ys}{r}\;\right)$$
chosen so as to turn the incircle of $\triangle A_\star B_\star C_\star$ into an origin-centered, axis-aligned ellipse with (horizontal) major radius $r$ and (vertical) minor radius $s$.
Let $A$, $B$, $C$ be the images of $A_\star$, $B_\star$, $C_\star$ under the transformation. One readily derives these relations:
$$\begin{align}
a^2 &= 12 \left( r^2 \sin^2\theta\phantom{\left(-60^\circ\right)}\; + s^2 \cos^2\theta \right) \\
b^2 &= 12 \left( r^2 \sin^2\left(\theta-60^\circ\right)+s^2 \cos^2\left(\theta-60^\circ\right) \right) \\
c^2 &= 12 \left( r^2 \sin^2\left(\theta+60^\circ\right)+s^2 \cos^2\left(\theta+60^\circ\right) \right) \\
\end{align}$$
Without too much trouble, one can derive 
$$r^2 = \frac{1}{36}\left(a^2 + b^2 + c^2 + 2 t^2 \right) \qquad s^2 = \frac{1}{36}\left(a^2+b^2+c^2-2t^2\right)$$
where 
$$t^4 = a^4 + b^4 + c^4 - b^2 c^2 - c^2 a^2 - a^2 b^2$$
and also
$$
\tan 2\theta = \frac{\sqrt{3}(c^2-b^2)}{-2a^2+b^2+c^2} \quad\to\quad
\cos 2\theta = \frac{-2a^2+b^2+c^2}{2t^2}
$$
Now, the inellipse's foci are the points $P_{\pm} := (\pm\sqrt{r^2-s^2},0)$, and we have (after some manipulation)
$$\begin{align}
|\overline{AP_{\pm}}|^2 &= \frac{1}{9}\left(-a^2+2b^2+2c^2+t^2\mp 12 r t \cos\theta \right)
\end{align}$$
I haven't found a particularly-insightful way to write the right-hand side. However, the product of the "$+$" and "$-$" forms gives a convenient difference of squares that reduces quite nicely to this:

$$|\overline{AP_{+}}||\overline{AP_{-}}| = \frac{1}{3}bc \quad\text{thus, also}\quad |\overline{BP_{+}}||\overline{BP_{-}}| = \frac{1}{3}ca \quad |\overline{CP_{+}}||\overline{CP_{-}}| = \frac{1}{3}ab \tag{$\star$}$$

Consequently, it's no wonder that
$$\frac{|\overline{AP_{+}}||\overline{AP_{-}}|}{|\overline{AB}||\overline{AC}|} + 
\frac{|\overline{BP_{+}}||\overline{BP_{-}}|}{|\overline{BC}||\overline{BA}|}+
\frac{|\overline{CP_{+}}||\overline{CP_{-}}|}{|\overline{CA}||\overline{CB}|} = 1$$
which is a somewhat weaker statement. In any case, $(\star)$ would seem to be the key to the relations you seek.
A: Sums  $$CA\cdot AB+BC\cdot AB+CA\cdot BC = s^2+(4R+r)r$$ and 
$$\frac{1}{CA\cdot AB}+\frac{1}{BC\cdot AB}+\frac{1}{CA\cdot BC}=\frac{1}{2rR}$$
are not related with Steiner inellipse. 
Two remaining sums are based on the products of distances from the foci $P$ and $Q$ of the Steiner inellipse of the triangle $\Delta ABC$ to its vertices, which can be found as follows. Let the vertices $A$, $B$, and $C$ and the foci $P$ and $Q$ have the complex coordinates $z_A$, $z_B$, $z_C$, $z_P$, and $z_Q$, respectively. Accoriding to Steiner’s Theorem [MP, Th. 2.1],  $z_P$ and $z_Q$ are given by the equality,
$$g\pm \sqrt{g^2-\frac f3},$$
where $g=\frac 13\left(z_A+z_B+z_C\right)$ is the centroid and $f=z_Az_B+ z_Bz_C+ z_Az_C$. 
Then, for instance,
$$|PA\cdot QA|=|z_P-z_A||z_Q-z_A|=\left|(g –z_A)^2- g^2+\frac f3\right|=
\left|2gz_A+z_A^2+\frac f3\right|.$$
I don’t know whether we can further simplify the expressions for two remaining sums. 
References 
[MP] D. Minda, S. Phelps Triangles, ellipses, and cubic polynomials, American Mathematical Monthly, 115 (8) (2008), 679-689.
