The crux of the proof is to indeed consider the Schur power matrix. Let $A$ be an $n\times n$ matrix. Then consider the $n! \times n!$ matrix $S(A)$ indexed by permutations $\sigma, \tau$ such that the $(\sigma,\tau)$-entry is given by $\prod_{i=1}^n a_{\sigma(i),\tau(i)}$ for $\sigma, \tau \in \mathfrak{S}_n$.
Schur's theorem. Let $A$ be a positive semidefinite matrix. Then, for any vector $x$, $x^TS(A)x \ge \det(A)x^Tx$.
It turns out that actually the determinant of $A$ is also an eigenvalue of $S(A)$ (and from the above theorem, it is the smallest eigenvalue). From this the Immanantal inequality also follows as a corollary. Let me repeat a proof of this claim below (this is not my original proof, obviously).
Proof. Consider the tensor product matrix $\otimes^n A$, and augment $x$ (which is a vector of dim $n!$) by padding with zeros appropriately to obtain a vector $z$ that is of dim $n^n$, so that $$x^TS(A)x= z^T(\otimes^n A)z.~~~~~~~~~~~~~~~(*)$$ Since $A$ is psd, we can write $A=B^TB$ for a lower-triangular matrix $B$. Thus, we have (using simple properties of the tensor product) $$[z^T(\otimes^n A)z = z^T(\otimes^n B^TB)z = z^T(\otimes^n B^T)(\otimes^n B)z.$$
Now by Cauchy-Schwarz we obtain \begin{align*} x^TS(A)x & \le \left[z^T(\otimes^n B^T)^2z\right]^{1/2}\left[z^T(\otimes^n B)^2z\right]^{1/2}\\ &= \left[z^T(\otimes^n (B^T)^2)z\right]^{1/2}\left[z^T(\otimes^n B^2)z\right]^{1/2}. \end{align*} Using the relation $(*)$ twice on the rhs above we thus obtain \begin{equation*} x^TS(A)x \le [x^TS(B^2)^Tx]^{1/2}[x^TS(B^2)x]^{1/2}. \end{equation*} Since $B^2$ is lower triangular, the diagonal entries of $S(B^2)$ are all easily seen to be actually equal to $\det(B^2)=\det(B)^2$. Moreover, $S(B^2)$ itself is lower-triangular, so that $x^TS(B^2)x = \det(B)^2x^Tx$ which equals $\det(A)x^Tx$. Substituting this in the final inequality above, the proof is complete.