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$DeclareMathOperatorImmImm$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 $mathfrak S_n$, $chi(e)det HleImm_chi(H)$, where the immanant $Imm_chi$ is defined by
$$Imm_chi(H):=sum_sigmachi(sigma)prod_i=1^nh_isigma(i).$$
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
$$K_rho:=sum_sigmaleft(prod_i=1^nh_isigma(i)right)rho(sigma).$$
It would be sufficient to prove that $Kge(det H)I_V$, 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
$$forall xiinmathbb C^frak S_n,,forall Hinmathbb H_n^+,qquad |xi|^2det Hlesum_sigma,thetabarxi_sigmaxi_thetaprod_ih_sigma(i)theta(i).$$
Is this inequality true?

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 $$|z^T(otimes^n (X^TY))z|^2 = |z^T(otimes^n X^T)(otimes^n Y)z|^2le z^T(otimes^n X^TX)z cdot z^T(otimes^n Y^TY)z$$
   


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 $|det C|^2=det(C^TC)=det(A)$.