Show $f(x,y,z) = \frac{\log(y/x)}{\log(z/x)}$ is quasi concave for $x \ge y > 0$, $x \ge z > 0$.

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?

• Could you clearly define the domain? Can $x/y/z$ all be negative? – LinAlg Mar 11 at 19:05
• @LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference. – Thomas Ahle Mar 11 at 19:50
• Ok, just to make sure, $\log(z/x)$ can be both negative and positive, right? Or did you mean $x \geq y > 0$ and $x > z > 0$? – LinAlg Mar 11 at 19:53
• @LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $\log(x/y)/\log(x/z)$. – Thomas Ahle Mar 11 at 19:56

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.
• 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? – Thomas Ahle Mar 11 at 21:16
• @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. – LinAlg Mar 12 at 0:44