$\DeclareMathOperator\Imm{Imm}$I am looking for a proof in English or French of Schur's theorem that, for every $H$ in the space $\mathbb H\_n^+$ of positive semi-definite Hermitian matrices, and every irreducible character $\chi$ of $\mathfrak S\_n$, $\chi(e)\det H\le\Imm\_\chi(H)$, where the *immanant* $\Imm\_\chi$ is defined by
$$\Imm\_\chi(H):=\sum\_\sigma\chi(\sigma)\prod\_{i=1}^nh\_{i\sigma(i)}.$$
Notice that the original paper [I. Schur, "Über endlicher Gruppen und Hermiteschen Formen" Math. Z., 1 (1918) pp. 184–207](https://link.springer.com/article/10.1007/BF01203611), is in German.


By the way, it seems that many authors relate Schur's theorem to symmetric polynomials. Is there any purely representation-theoretic proof of the inequality above? Let $(\rho,V)$ be a unitary representation whose character is $\chi$. We may associate to $\Imm\_\chi(H)$ a Hermitian matrix over $V$ by
$$K\_\rho:=\sum\_\sigma\left(\prod\_{i=1}^nh\_{i\sigma(i)}\right)\rho(\sigma).$$
It would be sufficient to prove that $K\ge(\det H)I\_V$, where $I\_V$ denotes the matrix of the scalar product. Because of Frobenius's theorem about the orthogonal decomposition of the regular representation, this amounts to proving that the analogous sum, where $\rho$ is replaced by the regular representation, satisfies the same estimate. In other words, Schur's theorem would be implied by the inequality
$$\forall \xi\in{\mathbb C}^{\frak S\_n},\,\forall H\in{\mathbb H}\_n^+,\qquad |\xi|^2\det H\le\sum\_{\sigma,\theta}\bar\xi\_\sigma\xi\_\theta\prod\_ih\_{\sigma(i)\theta(i)}.$$
Is this inequality true?



Schur's Theorem about immanants
Many thanks to **Denis** for pointing out my erroneous initial "proof". This time around the proof is correct, and directly proves the assertion in line 3 of the OP, i.e., $\chi(e)\det(A)\le d\_\chi(A)$ (I will write $d\_\chi(I)$ instead of $\chi(e)$ for uniformity). 


The explicit notation is cumbersome, so I am just writing a proof sketch. 


1. First, recall that $d\_\chi(A)=z^T(\otimes^n A)z$ for a suitable vector $z$
2. Next, use Cauchy-Schwarz to obtain $$|z^T(\otimes^n (X^TY))z|^2 = |z^T(\otimes^n X^T)(\otimes^n Y)z|^2\le z^T(\otimes^n X^TX)z \cdot z^T(\otimes^n Y^TY)z$$
3. Now write $A=C^TC$ for some upper triangular matrix $C$ (since $A$ is PSD we can do this). Then, put $X=C$ and $Y=I$ above, to obtain
4. $|z^T(\otimes^n C)z|^2 = |z^T(\otimes^n I)z|^2|\det C|^2 \le |z^T(\otimes^n C^TC)z|\cdot |z^T(\otimes^n I)z|$, where we used the upper triangular nature of $C$ for the first step. In other words, we have shown that
5. $d\_\chi(I)^2 \det(A) \le d\_\chi(A)d(I)$, since $|\det C|^2=\det(C^TC)=\det(A)$.


