Let $A=x+yi,\text{where} \left(y>0\right),B=-1,C=1\implies D=x-y+(x+y+1)i.$ We thus need to prove $$\sqrt{\left(x^2+y^2+1\right)^2-4x^2}+2\sqrt{x^2+y^2}\geq x^2+y^2+1+2y.$$After squaring, the latter writes as $$\sqrt{\left(x^2+y^2+1\right)^2-4x^2}\cdot\sqrt{x^2+y^2}\geq y(x^2+y^2+1).$$We square this too and get $\left(x^2+y^2+1\right)^2\geq4(x^2+y^2),$ which is $AM-GM.$ Equality at $x=0,$ i.e. $AB=AC.$ We also get equality at $x^2+y^2=1,$ which makes the angle from $A$ right. I have the strong belief we are talking about Tereshin here.

I am interested in another approaches, but especially I want to see the official solution.

I tried using Ptolemy's ineq but I couldn't do it, btw @above nice solution.

See here my second solution. I once created a problem with a part from this configuration. But it was a locus problem. I liked this one. Let $A=x+yi,\text{where} \left(y>0\right),B=-1,C=1\implies D=x-y+(x+y+1)i.$ We thus need to prove $$\sqrt{\left(x^2+y^2+1\right)^2-4x^2}+\sqrt{4x^2+4y^2}\geq x^2+y^2+1+2y.$$If $x=0$ or if $x^2+y^2=1,$ then we clearly have equality. Otherwise, since $$\left(\left(x^2+y^2+1\right)^2,4y^2\right) \text{strictly majorize} \left(\left(x^2+y^2+1\right)^2-4x^2,4x^2+4y^2\right),\text{then by Karamata}$$$$\sqrt{\left(x^2+y^2+1\right)^2-4x^2}+\sqrt{4x^2+4y^2}>\sqrt{\left(x^2+y^2+1\right)^2}+\sqrt{4y^2}= x^2+y^2+1+2y.$$The proof is complete. Equality if and only if $AB=AC$ or $AB\perp AC.$

i heard that there exists a " geometrical " solution

Let $E$ be a point such that $EA=AC$, $EA$ perpendicular to $AC$, and $B$ and $E$ on different sides with respect to $AC$. $AD=AB$, $AC=AE$, $\angle DAC = 90 + \angle BAC = \angle BAE$ $\Rightarrow$ $\bigtriangleup DAC \cong \bigtriangleup BAE$ $\Rightarrow CD=BE$...(1). Let $A'$ be the reflection of $A$ with respect to point $M$. From this, we have that $ABA'C$ is parallelogram $\Rightarrow AC=BA'$ and $\angle ABA' = 180 - \angle BAC$. $AD=AB$, $AE=AC=BA'$, $\angle DAE = 360 - \angle DAB - \angle BAC - \angle CAE = 360 - 90 - \angle BAC - 90 = 180 - \angle BAC = \angle ABA'$. $\Rightarrow \bigtriangleup DAE \cong \bigtriangleup ABA'$ $\Rightarrow DE= AA' = 2AM$...(2). From Pythagorean Theorem in $\bigtriangleup DAB$ and $\bigtriangleup EAC$, we can easily find that $DB=AB\cdot\sqrt{2}$ and $EC=AC\cdot\sqrt{2}$...(3). From here, we use Ptolemy's inequality for quadrilateral $DECB$ and we have that $DB\cdot CE + BC\cdot DE \ge DC\cdot BE$ ...(4). Substituting (1), (2), and (3) in (4), we have that $2AB\cdot AC + 2BC\cdot AM \ge CD^2$. The conclusion follows. Motivation to this solution: The LHS is symmetrical wrt to $B$ and $C$, so we try to take advantage of that.

Beautiful.

At equality, if we fix $B$ and $C$ then $A$ describes the union of a line and a circle.

Consider the point $E$, defined just like in post #7. Furthermore, let $X$ be the point where the A-Symmedian meets $(ABC)$. It is well known that $X$ lies on the A-Apollonian circle, so $XB/XC = AB/AC = AD/AE$. This, combined with the fact that $\angle{BXC} = \angle{DAE} = 180 - \angle{BAC}$, gives that $\triangle{BXC} \sim \triangle{DAE}$. Now the simple length relations $DE/BC = AD/BX = AB/BX = AM/MC = 2AM/BC$ give that $DE = 2AM$. Ptolemy on $BDEC$ finishes the problem since $BE = DC, BD = \sqrt{2}AB, CE = \sqrt{2}AC$.