= Solution
Work inside $V_{m+1}$, and take the <orthogonal complement for a bilinear form> of $U=V_m$ relative to this ambient space. If $z\in U\cap U^\perp$, then $\varphi(z,z)=0$, so positive definiteness on $U$ forces $z=0$.
Part (a)(iv) therefore gives $V_{m+1}=U\oplus U^\perp$, with $\dim U^\perp=1$. Choose $w\ne0$ spanning that complement and use the basis $e_1,\ldots,e_m,w$. Symmetry and orthogonality make its <Gram matrix>
$$
B=\begin{pmatrix}A_m&0\\0&d\end{pmatrix},\qquad d=\varphi(w,w).
$$
By part (b)(i), $\det B=(\det P)^2\det A_{m+1}>0$. Part (ii) applied to the already positive definite restriction gives $\det A_m>0$. Therefore
$$
d\det A_m=\det B>0,\qquad d>0.
$$
For a nonzero vector $u+tw$ with $u\in U$,
$$
\varphi(u+tw,u+tw)=\varphi(u,u)+t^2d>0.
$$
This proves \b[positive definiteness on $V_{m+1}$].
Finally $\det A_1>0$ says $\varphi(e_1,e_1)>0$, so the restriction to $V_1$ is positive definite. Induction using all successive positive leading <principal minors> reaches $V_n=V$. Combined with part (ii), this proves \b[<Sylvester's criterion>] in both directions.
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