Continuity of symmetric multilinear maps The following is Exercise 2.11-2 from Linear and Nonlinear Functional Analysis with Applications by Philippe G. Ciarlet.

Let $X$ and $Y$ be two normed vector spaces over the same field, and let $A$ be a multilinear, symmetric, and continuous mapping from $X\times X\times\cdots\times X$ ($k$ factors) into $Y$. Show that there exists a constant $C(k)$ such that $$\|A\|\leq C(k)\sup_{\|x\|\leq1}\|A(x,x,. . .,x)\|.$$

Here of course $\|A\|$ denotes the operator norm. The notation $C(k)$ means it's a constant that depends only on $k$ (and not on the particular $A$). The book says in a footnote that it has been proved that $k^k/k!$ is the best consant here.
I can't provide my attempts because I have no ideas yet... For example, I thought we could somehow write all $A(x_1,\ldots,x_n)$ as a sum of things like $A(x,x,. . .,x)$, but I don't know how to do that or whether it's actually correct. Any hints will be appreciated!
 A: Here is the proof for $k=3$. I do not know how to generalize this to larger $k$ (induction?). The idea is to observe the identity
$$
(x+y+z)^3 - (x+y-z)^3-(x-y+z)^3 + (x-y-z)^3 = 24xyz
$$
Take $x,y,z\in X$, set $v_1:=x+y+z$, $v_2 = x+y-z$, $v_3 = x-y+z$, $v_4 = x-y-z$. Then it holds
$$
24 A(x,y,z) = A(v_1,v_1,v_1) - A(v_2,v_2,v_2) + A(v_3,v_3,v_3) - A(v_4,v_4,v_4).
$$
Denote $M:=\sup_{\|x\|\le1} \|A(x,x,x)\|$. Then it follows
$$
\|A(x,y,z)\| \le 24^{-1} M ( \sum_{i=1}^4 \|v_i\|^3).
$$
Taking the supremum over $x,y,z$ with norm $\le 1$ on the left yields
$$
\|A\| \le 24^{-1}\cdot 4 \cdot 27M = \frac92 = \frac{3^3}{3!} M.
$$

In fact, this can be proven for all $k$. We have to use the following fact 
from Fischer https://www.jstor.org/stable/2690560 (in the notation from https://arxiv.org/pdf/1508.05235)
$$
2^{k-1} k! \prod_{i=1}^k x_i
= \sum_{I\subset \{2\dots k\}} (-1)^{|I|} (x_1 + \delta(I,2)x_2 + \dots + \delta(I,K))^k
$$
where
$$
\delta(I,j) = \begin{cases}-1 & \text{ if } j\in I,\\
1 & \text{ if } j\not\in I.\end{cases}
$$
To prove the identity, notice that $x_i=0$ is a zero of the right-hand side for all $i$. Using this identity, we get as above 
$$
2^{k-1} k! \|A(x_1,\dots, x_n)\| \le 2^{n-1} \ M\ k^k.
$$
