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I'd like to show that $$ \frac{\log\frac{y_1+y_2}{x_1+x_2}}{\log\frac{z_1+z_2}{x_1+x_2}} \ge \min\left\{ \frac{\log(y_1/x_1)}{\log(z_1/x_1)}, \frac{\log(y_2/x_2)}{\log(z_2/x_2)} \right\}. $$ This would follow if $f(x,y,z) = \frac{\log(y/x)}{\log(z/x)}$ was quasi concave.

One way to show this is if $f(x,y,z)\ge\alpha$ defines a convex set for all $\alpha\ge 0$. Plotting for various values of $\alpha$, this is clearly the case.

From the first order condition, it suffices if $f(\vec y) \le f(\vec x) \implies \nabla f(\vec x)^T(\vec y-\vec x) \ge 0.$ Let $x'\ge y',z'$ be such that $f(x',y',z')\le f(x,y,z)$ then we must show

$$z \log \left(\frac{x}{z}\right) (y x'-x y') \ge y \log \left(\frac{x}{y}\right) (z x'-x z'),$$ which unfortunately doesn't seem any easier.

Alternatively, since $\log x$ is quasi-linear, perhaps it suffices to show that $(x-y)/(x-z)$ is quasi-concave? I don't know a theorem to this effect, however.

Do you see a nice argument I might use to simplify this problem? Perhaps just tackling the original inequality directly?

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  • $\begingroup$ Could you clearly define the domain? Can $x/y/z$ all be negative? $\endgroup$ – LinAlg Mar 11 at 19:05
  • $\begingroup$ @LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference. $\endgroup$ – Thomas Ahle Mar 11 at 19:50
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    $\begingroup$ Ok, just to make sure, $\log(z/x)$ can be both negative and positive, right? Or did you mean $x \geq y > 0$ and $x > z > 0$? $\endgroup$ – LinAlg Mar 11 at 19:53
  • $\begingroup$ @LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $\log(x/y)/\log(x/z)$. $\endgroup$ – Thomas Ahle Mar 11 at 19:56
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Let me give you a possible method. Consider the superlevel set $\{(x,y,z) : \log(y/x) / \log(z/x) \geq \alpha, x \geq y > 0, x > z > 0 \}$. Since $\log(z/x)$ is negative, the sublevel set is equivalent to the set: $$\{(x,y,z) : \log(y/x) \leq \alpha \log(z/x), x \geq y > 0, x > z > 0 \}$$ $$=\{(x,y,z) : y \leq x^{1-\alpha}z^{\alpha}, x \geq y > 0, x > z > 0 \}.$$ Consider the Hessian of $f(x,z) = x^{1-\alpha}z^{\alpha}$. If it is negative semidefinite, the set is convex and you are done. Otherwise, try the same but with $x$ or $z$ on one side.

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  • $\begingroup$ The eigenvalues of the Hessian are $0$ and $-(1-\alpha) a x^{-\alpha-1} z^{\alpha-2} \left(x^2+z^2\right)$, so I guess it is negative semidefinite for all $\alpha\le1$. That suffices for my purposes! Is $\{(x,y,z) : y \le f(x,y)\}$ always convex when $f$ is concave? $\endgroup$ – Thomas Ahle Mar 11 at 21:16
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    $\begingroup$ @ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is $\{x : f(x) \leq 0\}$ where $f$ is convex. $\endgroup$ – LinAlg Mar 12 at 0:44

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