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"](http://mathworld.wolfram.com/CompletelyPositiveMatrix.html).



An inequality for positive definite matrices
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](http://mathworld.wolfram.com/CompletelyPositiveMatrix.html), 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](http://books.google.com/books?id=AllToPtYGGgC&lpg=PP1&dq=completely%20positive&pg=PP1#v=onepage&q=2.30&f=false), 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.$\qquad\blacksquare$


