In the page 10 of the book "The Malliavin Calculus and Related Topics" from Nualart one reads:
Letting $f=g$ in property $\rm (iii)$ obtains $$E(I_m(f)^2)=m!\|\tilde{f}\|^2_{L^2(T^m)}\leq m!\|f\|^2_{L^2(T^m)}.$$
The last inequality suggests that $\|\tilde f\|_{L^2(T^m)}\leq\| f\|_{L^2(T^m)}$ does not hold with equality,
otherwise one would write $$ \|\tilde f\|_{L^2(T^m)}=\| f\|_{L^2(T^m)}$$
Herre $\tilde f$ is the symmetrization of $f$.
The definition can be read in page 9
$\rm(ii)$ $I_m(f)=I_m(\tilde{f})$, where $\tilde f$ denotes the symmetrization of $f$, which means $$\tilde{f}(t_1,\ldots,t_m)=\frac1{m!}\sum_\sigma f(t_{\sigma(1)},\ldots,t_{\sigma(m)}),$$ $\sigma$ running over all permutations of $\{1,\ldots,m\}$,
Is there an example of when the inequality holds strictly?
Yes, take $f = 1_{[a,b]\times [c,d]}$. with $a > d$ so that the function does not intersect the diagonal.
The symmetrization of $f = \frac{1}{2}1_{[a,b]\times [c,d]} + \frac{1}{2}1_{[c,d]\times [a,b]}$
When we compute the $L^2$-norm, we find
$$\|\tilde f\|^2 = \frac{1}{4} \mu([a,b])\mu(c,d) + \frac{1}{4} \mu([c,d])\mu(a,b) < \mu([a,b])\mu(c,d) = \| f\|^2 $$