By the Cauchy-Schwarz Inequality,
$$\sqrt{PB^2 + PC^2} \sqrt{AC^2 + AB^2} \ge PB \cdot AC + PC \cdot AB.$$
Applying the (Generalized) Ptolemy's Inequality to quadrilateral $ABPC$ yields
$$PB \cdot AC + PC \cdot AB \geq PA \cdot BC.$$

Because $PA$ is the longest side of an obtuse triangle with side lengths $PA$, $PB$, $PC$, we have $PA > \sqrt{PB^2 + PC^2}$ and hence
$$PA \cdot BC \geq \sqrt{PB^2 + PC^2} \cdot BC.$$
Combining these three inequalities yields $\sqrt{AB^2 + AC^2} > BC$, implying that angle $BAC$ is acute.

Let $D$ and $Q$ be the feet of the perpendiculars from $B$ and $P$ to line $AC$, respectively. Then $DQ \leq BP$. Furthermore, the given conditions imply that $AP^2 > BP^2 + PC^2$, which can be written as $AP^2 - PC^2 > BP^2$. Hence,
$$\begin{align*} AQ^2 &\geq AQ^2 - QC^2 \\ &= (AP^2 - PQ^2) - (CP^2 - PQ^2) \\ &= AP^2 - PC^2 \\ &> BP^2 \\ &\geq DQ^2. \end{align*}$$
Let $\ell$ be the ray $AC$ minus the point $A$. Note that, since $PA > PC$, $Q$ lies on ray $\ell$. If $D$ did not lie on $\ell$, then $AQ$ would be less than or equal to $DQ$, a contradiction. Thus, $D$ lies on $\ell$, and angle $BAC$ is acute.

Set up a coordinate system on the plane with $A = (0,0)$, $B = (a,0)$, $C = (b,c)$, and $P = (x,y)$. Without loss of generality, we may assume that $a > 0$ and that $c > 0$. Proving that angle $BAC$ is acute is equivalent to proving that $b > 0$. Since $PA^2 > PB^2 + PC^2$,
$$x^2 + y^2 > (x-a)^2 + y^2 + (x-b)^2 + (y-c)^2.$$
Hence,
$$0 > (x-a)^2 - 2bx + b^2 + (y-c)^2 \geq -2bx.$$
Since $PA > PB$, we have $x > \frac{a}{2} > 0$. It follows that $b > 0$, as desired.

We first prove the following Lemma.

Lemma. For any four points $W$, $X$, $Y$, and $Z$ in the plane,
$$WY^2 + XZ^2 \leq WX^2 + XY^2 + YZ^2 + ZW^2.$$
Proof. Pick an arbitrary origin $O$ and let $w$, $x$, $y$, $z$ denote the vectors from $O$ to $W$, $X$, $Y$, $Z$, respectively. Then
$$\begin{align*} WX^2 + XY^2 + YZ^2 + ZW^2 - WY^2 - XZ^2 \\ &= |w-x|^2 + |x-y|^2 + |y-z|^2 + |z-w|^2 - |w-y|^2 - |x-z|^2 \\ &= w \cdot w + x \cdot x + y \cdot y + z \cdot z \\ &\quad -2(w \cdot x + x \cdot y + y \cdot z + z \cdot w - w \cdot y - x \cdot z) \\ &= |w+y-x-z|^2, \end{align*}$$
which is always nonnegative. Equality holds if and only if $w + y = x + z$, which is true if and only if $WXYZ$ is a (possibly degenerate) parallelogram.

Applying the Lemma to points $A$, $B$, $C$, and $P$ gives
$$\begin{align*} 0 \le AB^2 + BP^2 + PC^2 + CA^2 - AP^2 - BC^2 \\ &= (PB^2 + PC^2 - PA^2) + (AB^2 + AC^2 - BC^2) \\ &< 0 + (AB^2 + AC^2 - BC^2) \\ &= AB^2 + AC^2 - BC^2. \end{align*}$$
Therefore, angle $BAC$ is acute.

In this solution, $\sin^{-1}$ takes on values between $0^\circ$ and $90^\circ$. Note that $\angle PAB < 90^\circ$, since $PB < PA$. Applying the Law of Sines to triangle $PAB$ yields
$$\sin \angle PAB = \frac{PB}{PA} \sin \angle ABP \le \frac{PB}{PA}.$$
It follows that
$$\angle PAB \le \sin^{-1} \frac{PB}{PA}.$$
Since $PA^2 > PB^2 + PC^2$, we have similarly
$$\angle PAC \le \sin^{-1} \frac{PC}{PA} < \sin^{-1} \frac{\sqrt{PA^2 - PB^2}}{PA}.$$
Thus,
$$\begin{align*} \angle BAC &\le \angle BAP + \angle PAC \\ &< \sin^{-1} \frac{PB}{PA} + \sin^{-1} \frac{\sqrt{PA^2 - PB^2}}{PA}. \end{align*}$$
If $\theta = \sin^{-1} \frac{PB}{PA}$, then $\sin(90^\circ - \theta) = \cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{\sqrt{PA^2 - PB^2}}{PA}$.
Hence,
$$\angle BAC < \sin^{-1} \frac{PB}{PA} + \sin^{-1} \frac{\sqrt{PA^2 - PB^2}}{PA} = 90^\circ,$$
and angle $BAC$ is acute.

Note that $PA^2 > PB^2 + PC^2$. Regard $P$ as fixed and $A, B$, and $C$ as free to rotate on circles of radii $PA, PB$, and $PC$ about $P$, respectively. As $A, B, C$ vary, $\angle BAC$ will be maximized when $B$ and $C$ are on opposite sides of line $PA$ and $\angle ABP$ and $\angle ACP$ are right angles, i.e., lines $AB$ and $AC$ are tangent to the circles passing through $B$ and $C$.
Without loss of generality, we assume that $PA > PB \ge PC$. In this case, $ABPC$ is cyclic and $AB^2 = PA^2 - PB^2 > PC^2$, and similarly $AC^2 > PB^2$. Hence, on the circumcircle of $ABPC$, arcs $AB$ and $AC$ are bigger than arcs $PC$ and $PB$, respectively. Thus, $\angle BPC > \angle BAC$. Since these two angles are supplementary, angle $BAC$ is acute.