I am looking for a proof in English or French of Schur's theorem that for every $H\in{\mathbb H}_n^+$ (positive semi-definite Hermitian matrices) and every irreducible character $\chi$ of ${\frak S}_n$, $\chi(e)\det H\le{\rm Imm}_\chi(H)$, where the immanant is defined by $${\rm Imm}_\chi(H):=\sum_\sigma\chi(\sigma)\prod_{i=1}^nh_{i\sigma(i)}.$$ Notice that the original paper by I. Schur 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 $\chi$. With ${\rm Imm}_\chi(H)$, we may associate 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 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  ?

$\DeclareMathOperator\Imm{Imm}$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, 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?

I am looking for a proof in English or French of Schur's theorem that for every $H\in{\mathbb H}_n^+$ and every irreducible character $\chi$ of ${\frak S}_n$, $\chi(e)\det H\le{\rm Imm}_\chi(H)$, where the immanant is defined by $${\rm Imm}_\chi(H):=\sum_\sigma\chi(\sigma)\prod_{i=1}^nh_{i\sigma(i)}.$$ Notice that the original paper by I. Schur 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 $\chi$. With ${\rm Imm}_\chi(H)$, we may associate 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 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 ?

I am looking for a proof in English or French of Schur's theorem that for every $H\in{\mathbb H}_n^+$ (positive semi-definite Hermitian matrices) and every irreducible character $\chi$ of ${\frak S}_n$, $\chi(e)\det H\le{\rm Imm}_\chi(H)$, where the immanant is defined by $${\rm Imm}_\chi(H):=\sum_\sigma\chi(\sigma)\prod_{i=1}^nh_{i\sigma(i)}.$$ Notice that the original paper by I. Schur 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 $\chi$. With ${\rm Imm}_\chi(H)$, we may associate 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 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 ?