An inequality for positive definite matrices Let $K$ and $K^\prime$ positive definite $n \times n$ matrices, such that for all vectors $f \ge 0$ with nonnegative coordinates we have
$$\sum_{i,j} K_{ij} f_i f_j \le \sum_{ij} K^\prime_{ij} f_i f_j$$
Can someone prove the following inequality?
$$ \sum_{i,j} \exp K_{ij} \le \sum_{i,j} \exp K^\prime_{ij} $$
The reason why I believe it is true is that I have a proof using Slepian-Kahane-type comparison for Wick exponentials of Gaussian random vectors with covariance $K$ and $K^\prime$, which might look like a perverse way of doing something as simple-looking as this. Ideally, I'd like to see a more straightforward proof, or maybe some general suggestions on how to use that weird positivity condition on $K^\prime - K$.
Upd: The "weird positivity condition" is nothing but dual to something known by the name "complete positivity".
 A: Here is a simple proof (thanks to the OP for a proof that the Hadamard exponential is CP).
I'll write $K$ and $H \equiv K'$ to keep the notation simpler.
As $K \succeq 0$, it follows that its Schur exponent $[e^{k_{ij}}]$ is also psd. Thus, in particular, we can write the following factorization:
\begin{equation*}
 [e^{k_{ij}}] = \sum\nolimits_l u_l u_l^T.
\end{equation*}

Theorem. The Schur-Hadamard exponential $[e^{k_{ij}}]$ is completely positive, i.e., in the above factorization we can select the vectors $u_1,\ldots,u_n$ to be elementwise nonnegative. 

In addition to the OP's elegant argument (see comments below), another proof to this theorem may also be found as Theorem 2.30 (pp 131) in Completely Positive Matrices, by Berman and ‎Shaked-Monderer.
Then, consider
\begin{equation*}
 \sum\nolimits_{ij} e^{k_{ij}}(h_{ij}-k_{ij}) = \mbox{tr}[(H-K)\sum\nolimits_l u_lu_l^T] = \sum\nolimits_l u_l^T (H-K)u_l \ge 0,
\end{equation*}
where the final inequality follows from our hypothesis since $u_l \ge 0$.
Since $\exp(x)$ is a convex function, we have
\begin{equation*}
 e^h \ge e^k + e^k(h-k),\qquad\forall h, k.
\end{equation*}
Thus, it follows that
\begin{equation*}
  \sum_{ij} e^{h_{ij}} \ge \sum_{ij} e^{k_{ij}} + \sum_{ij} e^{k_{ij}}(h_{ij}-k_{ij}).
\end{equation*}
But the above argumentation showed that the last term on the rhs is nonnegative.$\qquad\blacksquare$
