# Demonstrate that for a concave downward function that the value of the derivative $\frac{dy}{dx}$ is always higher than $\frac{y}{x}$

I would like to demonstrate that for a concave downward function passing through the origin that the value of the derivative $$\frac{dy}{dx}$$ is always higher than $$\frac{y}{x}$$. As shown in figure, it is simple to understand graphically, but for a generical concave downward function, without having a specific function, I don't know how to start.

• the graph shows that the inequality is reversed Nov 15, 2021 at 17:54
• The thing you are trying to prove is not true Nov 15, 2021 at 18:00
• If you assume that the function passes through the origin then you should state so in the question.
– dxiv
Nov 15, 2021 at 18:30
• Sorry @GCab I wrote it badly in the figure, Now I fixed it. Nov 15, 2021 at 19:38
• @dxiv sorry I forgot to mention it. Nov 15, 2021 at 19:46

Let's use $$y=-(x-2)^2+C$$ Then at $$x=1$$ one gets $$\frac yx=C-1$$ and $$\frac{dy}{dx}=-2(1-2)=2$$ Notice that you can change the $$y/x$$ ratio to any number, while the derivative is always constant. So you can't prove that it is smaller or larger.
The definition of concave function $$f$$ is that $$\forall \lambda\in(0,1)$$ $$f((1-\lambda)a+\lambda b)\ge(1-\lambda)f(a)+\lambda f(b)$$ So let's choose $$a=0$$ and $$b=x$$. We know that $$f(0)=0$$, so $$f(\lambda x)\ge\lambda f(x)$$ Then we want to calculate $$\frac{f(x)-f(x-h)}h$$ We can choose $$x-h=\lambda x$$ or $$h=(1-\lambda)x$$. $$\frac{f(x)-f(x-h)}h=\frac{f(x)-f(\lambda x)}{(1-\lambda)x}\le\frac{f(x)-\lambda f(x)}{(1-\lambda)x}=\frac{f(x)}{x}$$
• I used $f(0)=0$ to get from definition to the second equation in the edited part. If my final formula is true for every $\alpha$, it's true in the limit as well Nov 16, 2021 at 13:11
• which parameter is $\alpha$? This demonstration is true also if I choose different values of a and b? Nov 16, 2021 at 13:49