Nuclear norm as minimum of Frobenius norm product

Here's a nice to way to go about this. I will write $\|\cdot\|_1$ for the nuclear norm, and $\|\cdot\|_2$ for the Frobenius norm.

First, we have the matrix Hölder inequality, which implies $\|UV\|_1\leq\|U\|_2\|V\|_2$. We also have $\|U\|_2\|V\|_2 \leq \tfrac{1}{2}(\|U\|_2^2 + \|V\|_2^2)$. Taken together, these give $$ \|X\|_1 \leq \min_{UV = X} \|U\|_2 \|V\|_2 \leq \min_{UV = X} \tfrac{1}{2}(\|U\|_2^2 + \|V\|_2^2). $$ To see that both inequalities are tight, let $X = S(X^*X)^{1/2}$ be the polar decomposition of $X$, with a partial isometry $S$ such that $S^*S$ is the support projection of $X^*X$. Taking $U = S(X^*X)^{1/4}$ and $V = (X^*X)^{1/4}$ works, since $$ \|S(X^*X)^{1/4}\|_2^2 = \|(X^*X)^{1/4}\|_2^2 = \mathrm{tr}((X^*X)^{1/2}) = \|X\|_1, $$ where the first step also uses that $S^*S$ is the support projection of $X^*X$.

One of the advantages of this more abstract argument over a brute-force calculation is that this argument also applies in any von Neumann algebras equipped with a normal semifinite trace. Therefore the equations are still valid in that context.