Finding a Lower Bound of a Function Let
$$
f(x, y) = \frac{x + y}{2e}
$$
and
$$
g(x, y) = \frac{(x+y)+\sqrt{x^2 + y^2 + 6xy}}{2e^{\frac{\sqrt{x^2 + y^2 + 6xy}}{2(x + y)}}} 
$$
I am looking for the largest $K$ so that,
$$ \forall x, y > 0, \frac{g(x, y)}{f(x, y)} > K$$
By numerical computation I found that $K$ must be around $3.3$. But the best thing I could prove is about 2.28 for $K$ using the following method,
$$
x^2 + y^2 + 2xy < x^2 + y^2 + 6xy < (\sqrt{3}x)^2 + (\sqrt{3}y)^2 + 2(\sqrt{3}x)(\sqrt{3}y)
$$
Which results in,
$$
(x+y) < \sqrt{x^2 + y^2 + 6xy} < \sqrt{3}(x+y)
$$
So,
$$
\frac{g(x, y)}{f(x, y)} > \frac{\frac{x+y}{e^{\sqrt{3}/2}}}{\frac{x+y}{2e}} > 2.28
$$
I'm wondered if anyone can propose a way to prove a larger K?
Note: This is the improvement of the network throughput based on the method I proposed (g(.)) over an old method (f(.)). So, this is important for me to show that how much better my proposed method works.
Thanks in advance :)
 A: $K=(1+\sqrt{2})\mathrm{e}^{1-\sqrt{2}/2}=3.23576963$.
In short, this is because $x^2+y^2+6xy\le2(x+y)^2$.
For a more detailed proof, write the ratio $g(x,y)/f(x,y)$ as $h(z)=\mathrm{e}^{1-z/2}(1+z)$ with $z^2=(x^2+y^2+6xy)/(x+y)^2$. Now, $z\mapsto h(z)$ is a decreasing function on $z\ge1$ (differentiate the logarithm) and, for every positive $x$ and $y$, $z^2-1=4xy/(x+y)^2\le1$ hence $1\le z^2\le 2$. This means that $h(z)$ is minimal when $z^2=2$, that is, when $x=y$. This yields the value of $K$ given above, if one replaces the strict inequality sign $>$ in your post by a $\ge$ sign.
A: The minimum of $g(x,y)/f(x,y)$ occurs at the points $(x,y)$ where $x=y$, and is given by
$$
K = (1 + \sqrt 2 )e^{1 - \sqrt 2 /2} \approx 3.235769626477154.
$$ 
Sketch of proof. Using the transformations $s=x+y$, $t=xy$, one finds that
$$
\frac{{g(x,y)}}{{f(x,y)}} = e\frac{{1 + \xi }}{{e^{\xi /2} }},
$$
where
$$
\xi  = \sqrt {1 + \frac{{4xy}}{{(x + y)^2 }}} .
$$
The maximum of $\xi$ occurs at the points $(x,y)$ where $x=y$ (straightforward exercise), hence equal to $\sqrt{2}$.
The result then follows from the fact that the function $(1 + \xi )/e^{\xi /2}$
is decreasing (which is verified upon differentiation).
EDIT: It is interesting (and maybe important) to note that the function $g(x,y)/f(x,y)$, $x,y > 0$, though quite complicated in appearance, is not far from being a constant. Indeed, by considering the infimum of $\xi$ (which is equal to $1$), we get the (tight) upper bound of $2\sqrt{e} \approx 3.297442541400256$, so
$$
3.2357 < (1 + \sqrt 2 )e^{1 - 2/\sqrt 2 }  \le \frac{{g(x,y)}}{{f(x,y)}} < 2\sqrt e  < 3.2975,
$$
for any $x, y > 0$.
