Lipschitz constant for map between triangles Let $T_1$ and $T_2$ be any two euclidean triangles with labeled sides. The sides are labeled respectively $e_1^1,e_2^1,e_3^1$ and $e_1^2,e_2^2,e_3^2$. Call $A:T_1\rightarrow T_2$ the affine map which sends sides to corresponding sides, i.e. $e_i^1\mapsto e_i^2$, $i=1,2,3$.
Call $\mathcal{F}$ the set of differentiable maps which send $T_1$ to $T_2$ and sides to corresponding sides, clearly $A\in \mathcal{F}$. For every $f\in \mathcal{F}$ define $L(f):=\max\{Lip(f),Lip(f^{-1})\}$ where $Lip(f)$ is the Lipschitz constant of $f$.
Question(s): Is it true $L(A)=\inf\limits_{f\in\mathcal{F}}L(f)$? If not, is there a map which realizes the infimum? Which one?
Suppose $L(A)=Lip(A)$, then the answer to the first question is yes if $Lip(A)$ is obtained along a side $e$ of $T_1$: $A(e)/e=Lip(A)$, but for every other $f\in\mathcal{F}$ it's true $L(f)\ge f(e)/e\ge A(e)/e=L(A)$.
Unfortunately in case $L(A)$ is not obtained along a side of a triangle the previous inequality can not be used and I don't know how to proceed.
Additional question: what are the answers to the previous questions in case one or two sides of $T_1$ are mapped linearly? 
 A: Consider the example in the comment above, i.e. $A=(-1,0)$, $B=(1,0)$, $C=(0,1)$, $A'=A=(-1,0)$, $B'=B=(1,0)$ and $C'=(3,0)$. Then the affine map mapping $A,B$ and $C$ to $A'$, $B'$ and $C'$ respectively is given by
$$
x \mapsto Ax \text{ with } A=\begin{pmatrix} 1 & 2\\0 & 0\end{pmatrix}.
$$
The Lipschitz constant is given by the maximal singular value of $A$ which is $\sqrt{5}$. Introduce now the points $D=(0,0)$ and $D'=(\frac{1}{2},0)$. Then the matrices of the affine maps from $ADC$ to $A'D'C'$ and from $CDB$ to $C'D'B'$ are
$$
\begin{pmatrix} \frac{3}{2} & \frac{3}{2}\\0 & 0\end{pmatrix},\quad \begin{pmatrix} \frac{1}{2} & \frac{3}{2}\\0 & 0\end{pmatrix}.
$$
The corresponding Lipschitz constants are $\sqrt{2}\frac{3}{2}=\frac{3}{\sqrt{2}}<\sqrt{5}$ and $\sqrt{\frac{5}{2}}<\sqrt{5}$. Hence we see that the new map has a smaller Lipschitz constant. The new Lipschitz constant is equal to the maximal ratio of the sides of the triangles and can therefore not be improved.
A: EDIT: My first answer falsely claimed a positive proof (the mistake was pointed out by user372511). Here is instead an explicit counterexample.
Consider the following triangles with $T_1 = ABC$ and $T_2=ABC'$

The following map 
$$
f : \begin{pmatrix} x \\ y \end{pmatrix} \rightarrow  \begin{pmatrix} 1+2^y -2^{1-x} \\ 1+2^y-2^{1-y}\end{pmatrix}
$$
maps $T_1$ to $T_2$ and has a smaller Lipschitz constant than the affine map.
Indeed, the affine map is given by the matrix $A=\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} $. The maximal singular value gives the Lipschitz constant
$$ \mathrm{Lip}(A) = \sqrt{3+\sqrt{5}} \sim 2.288 $$
Let $J_f$ be the Jacobian of $f$. With the help of Mathematica, one can find the maximal singular value of $J_f(x,y)$ in the triangle. It is in fact reached at $(0,0)$ and gives the Lipschitz constant of $f$ in the triangle
$$ \mathrm{Lip}(f) = \sqrt{7+\sqrt{13}} \ln(2) \sim 2.257$$
