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 :)