How to prove that triangle inscribed in another triangle (were both have one shared side) have lower perimeter?

One way to see that this must be true is to look at the ellipse through $B$ with foci $A$ and $C$. The interior of this ellipse is precisely all those point $P$ such that $|AP| + |PC| < |AB| + |BC|$. And if $P$ is inside the triangle, then it is inside the ellipse too.

This is a special case of "Archimedes' axiom", that if one convex curve $\gamma_1$ is inside another ($\gamma_2$), then $\gamma_1$ is shorter than $\gamma_2.$ Archimedes needed this to justify his computing the perimeter of a circle by inscribed/circumscribed polygons, and could not prove it, so made it an axiom. In general, this is a nontrivial fact, in this case it is a tedious computation (if you don't want to use the general machine).

EDIT the most elegant proof is via Crofton's formula, which says that the length of a convex curve is equal (up to normalizing constant) to the measure of the lines which intersect the interior -- that measure is obviously monotonic under containment...

Since $P$ is inside $\triangle ABC$, the convex hull span by the three vertices $A$, $B$, $C$, there exists 3 numbers $\alpha, \beta, \gamma \ge 0$ such that

$$\alpha+\beta+\gamma = 1\quad\text{ and }\quad \vec{P} = \alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}$$

This implies $$ |AP| = |(1-\alpha)\vec{A} - (1-\alpha-\gamma)\vec{B} - \gamma\vec{C}| = |(1-\alpha)(\vec{A}-\vec{B}) + \gamma (\vec{B}-\vec{C})|\\ \le (1-\alpha) |AB| + \gamma |BC| $$ and $$ |CP| = |-\alpha \vec{A} - (1-\alpha-\gamma)\vec{B} + (1-\gamma)\vec{C}| = |\alpha (\vec{B}-\vec{A}) + (1-\gamma)(\vec{C}-\vec{B})|\\ \le \alpha |AB| + (1-\gamma)|BC| $$ Summing these two inequalities immediately gives us $|AP| + |CP| \le |AB| + |BC|$.

Since numbers like $\alpha$ are ratios of distance of $P$ to $BC$ versus that of $A$ to $BC$, it is easy to translate above vector based inequalities to a pure geometric proof.

Construct a line through $P$ parallel to $BC$ and let it intersect $AB$ at $D$. Construct another line through $P$ parallel to $AB$ and let it intersect $BC$ at $E$. It is easy to see $|BE| = |PD|$ and $|BD| = |EP|$. Apply triangle inequalities to $\triangle APD$ and $\triangle CPE$, we have

$$|AP| + |CP| \le \Big(|AD| + |PD|\Big) + \Big(|CE| + |EP|\Big) = \Big(|AD| + |BE|\Big) + \Big(|CE| + |BD|\Big)\\ = \Big(|AD| + |DB|\Big) + \Big(|CE| + |BE|\Big) = |AB| + |BC|$$