We sort $P_1,\cdots,P_n$ such that $$\angle BAP_1 < \angle BAP_2 < \cdots < \angle BAP_n$$ We first use $n+1$ triangles: $\triangle BAP_1$, $\triangle P_jAP_{j+1}$ for $1\le j\le n-1$, and $\triangle P_nAC$. Let $\theta_j=\angle P_jBC$. Let $m$ be the length of the longest increasing sequence of $\theta_j$'s, and let $k$ be the length of the longest decreasing sequence of $\theta_j$'s. Claim: $mk\ge n$. Proof: This can be clear with a Young Tableaux. We can also partition a permutation into at most $m$ decreasing sequences by using posets (and using Dilworth's theorem): let $v_j > v_k$ if $j<k, \theta_j>\theta_k$. Then an increasing sequence is an antichain and a decreasing sequence is a chain, and by Dilworth's theorem, we can partition a poset $\{v_1,\cdots,v_n\}$ into $m$ chains, where $m$ is the number of antichains. If $k\ge \sqrt{n}$, then if $\theta_{x_1} > \cdots > \theta_{x_k}$, we can use triangles $\triangle P_{x_{j+1}}P_{x_j}B$ for $j=1,\cdots,k-1$ then tile $\triangle P_{x_k}BC$. This allows us to tile $k$ triangles. Otherwise, $m\ge \sqrt{n}$. Lemma: In a triangle $\triangle ABC$, if $P,Q$ satisfy $\angle PBC < \angle QBC, \angle PAB < \angle QAB$, then we must have $\angle PCA > \angle QCA$. Proof: Note $$\prod\limits_{cyc} \frac{\sin PAC}{\sin PAB} = \prod\limits_{cyc} \frac{PC \sin \angle APC}{AC} / \frac{PB \sin \angle APB}{AB} = 1$$ Similarly, $$\prod\limits_{cyc} \frac{\sin QAC}{\sin QAB} =1$$ However, if we write this as $$\prod\limits_{cyc} \frac{\sin PAC}{\sin PCA} = \prod\limits_{cyc} \frac{\sin QAC}{\sin QCA}$$ We can see that $\sin(x-t)/\sin(t)$ is decreasing as $t$ increases, so we cannot have $\angle PBC < \angle QBC, \angle PAB < \angle QAB$ and $\angle PCA \le \angle QCA$. This implies that if $\theta_{b_1} < \cdots <\theta_{b_m}$, then $\angle P_{b_1}CB > \angle P_{b_2}CB > \cdots > \angle P_{b_m}CB$, so again we have placed at least $\sqrt{n}$ triangles.

Call a triangle "good", if it contains at least one of $A,B,C$ Main claim: By Erdos-Szekeres we can find $k:=\lceil\sqrt{n}\rceil$ points arranged like this: Thus we can first partition $\triangle ABC $ to: There are $2k+1$ good triangles now and every other point adds at least one good triangle so there are at least $$(2k+1)+(n-k)=n+k+1\ge n+\sqrt{n}+1$$good triangles. $\blacksquare$