Equivalence of two inequalities related to the Grothendieck inequality Let $(a_{ij})$ be a real $m\times n$ matrix and $H$ be a Hilbert space. I wonder how to see the following two statements are equivalent. 
(1) For any vectors $u_i,v_j$ in $H$ of norm 1, we have 
$$\left|\sum_{i,j}a_{i,j}\langle u_i,v_j \rangle \right|\le K,$$
where $K$ is a constant.
(2) For any vectors $u_i,v_j$ in $H$, we have 
$$\left|\sum_{i,j}a_{i,j}\langle u_i,v_j \rangle \right|\le K \max_i \|u_i\|\max_j \|v_j\|,$$
where $K$ is a constant.
Clearly (2) implies (1). For the implication $(1)\Rightarrow (2)$, I get stuck since individual terms may be positive or negative and I have trouble when I do the amplification. 
Source: Vershynin HDP book page 61
Plus, you might have an urge to add $\pm 1$ to $u_i,v_j$ to make every term positive, but I don't think one can do that (so that the absolute value can be removed), because the matrix formed by $u_iv_j$ has at most rank 1. Let me know if I am wrong.
 A: It seems that the goal is to prove the following: suppose that that the following version of Theorem 3.5.1 is proved (using (a) of Exercise 3.5.2.):

Theorem A Suppose that for all $x_i,y_j$, $\lvert \sum_{i,j}a_{i,j}x_iy_j\rvert \leqslant \max_i \lvert x_i\rvert\max_j \lvert y_j\rvert$.
Then for all Hilbert space $H$ and all unit vectors $u_i$ and $v_j$,  $\lvert  \sum_{i,j}a_{i,j}\langle u_i,v_j\rangle\rvert\leqslant K$, where $K$ is an absolute constant.

Then the following Theorem B can be proven:

Theorem B Suppose that for all $x_i,y_j$, $\lvert \sum_{i,j}a_{i,j}x_iy_j\rvert\leqslant \max_i \lvert x_i\rvert\max_j \lvert y_j\rvert$.
Then for all Hilbert space $H$ and all vectors $u_i$ and $v_j$,  $\lvert  \sum_{i,j}a_{i,j}\langle u_i,v_j\rangle\rvert\leqslant K\max_i\lVert u_i\rVert \max_j\lVert v_j\rVert $, where $K$ is an absolute constant.

In other words, we have to prove that Theorem A implies the apparently stronger Theorem B. To do so, let $(a_{i,j})$ be such that for all $x_i,y_j\in \{-1,1\}$, $\lvert \sum_{i,j}a_{i,j}x_iy_j\rvert\leqslant 1$. Let $H$ be a Hilbert space and $u_i$, $v_j$ be not necessarily unit vectors. Assume without loss of generality that none of the $u_i$ and $v_j$ is zero. Denote $U=\max_i\lVert u_i\rVert$ and $V :=\max_j\lVert v_j\rVert $ and define
$$
b_{i,j}:=a_{i,j} \frac{\lVert u_i\rVert}U\frac{\lVert v_j\rVert}V.
$$
Let $x_i,y_j$ be real numbers. Then
$$
\left\lvert \sum_{i,j}b_{i,j}x_iy_j\right\rvert= \left\lvert \sum_{i,j}a_{i,j} \frac{\lVert u_i\rVert}Ux_i\frac{\lVert v_j\rVert}Vy_j\right\rvert
\leqslant \max_i\frac{\lVert u_i\rVert}U\lvert x_i\rvert\frac{\lVert v_j\rVert}V\lvert y_j\rvert\leqslant \max_i \lvert x_i\rvert \lvert y_j\rvert $$
hence $(b_{i,j})$ satisfies the assumption of Theorem A.
Applying it to $u'_i:= u_i/\lVert u_i\rVert$ and $v'_j:= v_j/\lVert v_j\rVert$, we reach the wanted conclusion.
