Projective norm for Banach spaces The projective norm of tensors from (algebraic) tensor product of Banach spaces $X,Y$ is defined as
$$\|t\|_\wedge = \inf\left\{ \sum\limits_{j=1}^N \|x_j\|\|y_j\| \, : \, t=\sum\limits_{j=1}^N x_j\otimes y_j\right\}.$$
In one place I found slightly different definition:
$$\|t\|_\wedge '=\inf\left\{\left(\sum\limits_{j=1}^N \|x_j\|^2\right)^{1/2}\left(\sum\limits_{j=1}^N \|y_j\|^2\right)^{1/2} \, : \, t=\sum\limits_{j=1}^N x_j\otimes y_j\right\}.$$
Are they indeed equal? It is easy to see one inequality, but why we have the opposite one?

The two norms are indeed equal. On the one hand, by the Cauchy-Schwarz inequality we have
$$\sum_{j = 1}^{N} \lVert x_j\rVert \lVert y_j\rVert \leqslant \Biggl(\sum_{j = 1}^{N} \lVert x_j\rVert^2\Biggr)^{1/2} \Biggl(\sum_{j = 1}^{N} \lVert y_j\rVert^2\Biggr)^{1/2}$$
which yields $\lVert t\rVert_{\wedge} \leqslant \lVert t\rVert'_{\wedge}$. I suppose that's the inequality you saw.
On the other hand, consider when we have equality in the Cauchy-Schwarz inequality. A particular case where we have equality is when $\lVert x_j\rVert = \lVert y_j\rVert$ for $1 \leqslant j \leqslant N$.
We can always achieve that by moving a scalar factor from $x_j$ to $y_j$, unless one of $x_j$ and $y_j$ is zero. But excluding all representations
$$t = \sum_{j = 1}^{N} x_j \otimes y_j \tag{1}$$
where at least one of the $x_j$ or $y_j$ is zero doesn't change either norm. For the first, including or excluding such terms doesn't change $\sum \lVert x_j\rVert \lVert y_j\rVert$ at all, for the second, the value for the sum without zero factors is smaller (or equal, if $x_j$ and $y_j$ are both zero for such terms), so the terms we exclude can't make the infimum, i.e. $\lVert t\rVert'_{\wedge}$ smaller.
Thus for a representation $(1)$ where no $x_j$ or $y_j$ vanishes, define
$$c_j = \sqrt{\frac{\lVert x_j\rVert}{\lVert y_j\rVert}}$$
and $x_j' = c_j^{-1}\cdot x_j$, $y_j' = c_j \cdot y_j$. Then we have
$$t = \sum_{j = 1}^{N} x_j' \otimes y_j'$$
and
$$\sum_{j = 1}^{N} \lVert x_j\rVert \lVert y_j\rVert = \sum_{j = 1}^{N} \lVert x_j'\rVert \lVert y_j'\rVert = \Biggl(\sum_{j = 1}^{N} \lVert x_j'\rVert^2\Biggr)^{1/2} \Biggl(\sum_{j = 1}^{N} \lVert y_j'\rVert^2\Biggr)^{1/2}$$
which shows
$$\lVert t\rVert'_{\wedge} \leqslant \lVert t\rVert_{\wedge}\,.$$
