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Let $$ f(x, y) = \frac{x + y}{2e} $$


$$ g(x, y) = \frac{(x+y)+\sqrt{x^2 + y^2 + 6xy}}{2e^{\frac{\sqrt{x^2 + y^2 + 6xy}}{2(x + y)}}} $$

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,

$$ (x+y) < \sqrt{x^2 + y^2 + 6xy} < \sqrt{3}(x+y) $$


$$ \frac{g(x, y)}{f(x, y)} > \frac{\frac{x+y}{e^{\sqrt{3}/2}}}{\frac{x+y}{2e}} > 2.28 $$

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 :)

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up vote 6 down vote accepted


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.

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You are my hero man :D thanks a lot. – Helium Mar 16 '11 at 23:37

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 $x, y > 0$.

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Wow, it must be true (based on my numerical result), thanks a lot. – Helium Mar 16 '11 at 23:10
Thanks Shai, your claim is quite the same as the one provided by Didier but your description looks a bit simpler to understand. PS: what happened to $s$ and $t$? – Helium Mar 16 '11 at 23:46
Moshen, this is only a sketch of proof... Another step, which I haven't mentioned, is to use $t=us^2$. – Shai Covo Mar 16 '11 at 23:57
@Shai Your proof is complete, all you have to do is to get rid of your $s$, $t$ and $u$ and to work with your $\xi$ (which is equivalent to my $z$). – Did Mar 17 '11 at 8:30
@Shai: Your EDIT is both interesting and important to me. Thanks for your point. – Helium Mar 17 '11 at 10:48

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