See here for picture

Very nice!

Perform inversion in $M$. Now we need to prove that for any point $M$ inside a parallelogram $ABCD$ next inequality holds: $$AM \cdot CM + BM \cdot DM \geq AB \cdot BC$$Draw line through $M$ parallel to $AB$. Move $AM, DM$ to the top, $BM, CM$ to the bottom using parralel moving such that $AM \rightarrow XY, DM \rightarrow YZ, BM \rightarrow TX, CM \rightarrow ZT; X \in AB $ etc. Now for quarilateral $XYZT$ we need to prove that $$XY \cdot ZT + YZ \cdot TX \geq XZ \cdot YT$$which is Ptolemeus inequality

It's not hard to see that the angle condition implies that $(AMD),(BMC)$ and $(AMB),(CMD)$ are pairs of tangent circles. Therefore, after inverting at $M$ with power $r^2$ we see that we get $A^*B^*C^*D^*$ a parallelogram. After using the inversion distance formula, we see that the inequality is equivalent to showing \[MA^*\cdot MC^*+MB^*\cdot MD^*\ge xy\]where $x=A^*B^*=C^*D^*$ and $y=A^*D^*B^*C^*$. We'll drop the stars now. Let $M'=M+\vec{CB}$. By Ptolemy on $M'AMB$, we see that \[M'M\cdot AB\le M'A\cdot MB+MA\cdot M'B,\]or \[xy\le MD\cdot MB+MA\cdot MC,\]as desired.

Invert about $M$ with arbitrary power $R^2$. Briefly dropping stars, the new angle conditions are $\angle AMB=\angle MAD+\angle MBC$ and $\angle AMD=\angle MAB+\angle MDC$. From the second condition, $\angle MAD+\angle MDA +\angle MAB+\angle MDC=180$, so $\angle BAD+\angle CDA=180$. Similarly, by the first condition, $\angle ABC+\angle BCD=180$. Therefore, $A^*B^*C^*D^*$ is a parallelogram. We have $MA=\tfrac{R^2}{MA^*}$, and so on. We can verify that $AB = \frac{R^2}{MA^*\cdot MB^*} \cdot A^*B^*$ by using the Law of Sines. The LHS is \[ AM\cdot CM + BM\cdot DM = \frac{R^4}{\prod_{\text{cyc}} MA^*} [MA^*\cdot MC^* + MB^*\cdot MD^*].\]And the RHS is \[ \sqrt{\prod_{\text{cyc}} AB} = \frac{R^4}{\prod_{\text{cyc}} MA^*} \cdot \sqrt{\prod_{\text{cyc}} A^*B^*} = \frac{R^4}{\prod_{\text{cyc}} MA^*} \cdot A^*B^*\cdot B^*C^*\]since $A^*B^*C^*D^*$ is a parllelogram. So we want to show \[ MA^*\cdot MC^* + MB^*\cdot MD^* \ge A^*B^*\cdot B^*C^*. \]The following finish is quite nice. Drop all stars. Construct $M'$ which is the point such that $MM'CD$ is a parallelogram; hence $MM'BA$ is also a parallelogram. Ptolemy on $BM'CM$ gives \begin{align*} AB\cdot BC &= MM'\cdot BC \\ &\le CM'\cdot MB + BM'\cdot MC \\ &= MD\cdot MB + MA\cdot MC. \end{align*}This completes the proof.

Please, I do not understand the "New angle conditions...

Invert about $M$. Since I'm lazy we omit stars. Thus we have the angle conditions \begin{align*} \angle AMB &= \angle DAM + \angle CBM \\ \angle AMD &= \angle BAM + \angle CDM. \end{align*}The first condition implies $\overline{BC} \parallel \overline{AD}$ (just draw a diagram!) and similarly, $ABCD$ is a parallelogram. Now consider $M'$ such that $\triangle BMC \cong \triangle AMD$. By Ptolemy inequality, $$AB \cdot BC = MM' \cdot BC \leq CM \cdot AM + BM \cdot DM.$$Using the inversion distance formula yields the desired result.

We consider inversion about $M$. We can see that $\angle MA^*D^* + \angle MB^*C^* = \angle A^*MB^*$. Thus, if we consider a line through $M$ parallel to $A^*D^*$ then we can see that $A^*D^* \parallel B^*C^*$ and similarly for the other pair of lines. Thus, $A^*B^*C^*D^*$ is a parallelogram. Now we need to prove that for any point in a parallelogram $M$ then we have \[AM \cdot CM + BM \cdot DM \geq AB \cdot BC\]Now consider a point $M'$ sucht that $MM'CD$ is a parallelogram. Then we can use inversion distance formula to yield the result. $\blacksquare$

The angle conditions imply $(MAB)$ and $(MCD)$ are externally tangent, as are $(MBC)$ and $(MDA)$. Inverting at $M$ with radius 1 makes $ABCD$ a parallelogram, and inversion distance formula rewrites the desired as \[AM \cdot CM + BM \cdot DM \ge \sqrt{AB \cdot BC \cdot CD \cdot DA} = AB \cdot BC.\] Translating $\triangle MCD$ to $\triangle M'C'D'$ such that $C' = B$ and $D' = A$, this inequality is simply Ptolemy on $M'AMB$. $\blacksquare$

another problem where I missed the finish ;-;