Show that for $f(x)=x^\alpha$, $f'(x+\frac{1}{2}a) - \frac{1}{2}f'(x+a) - \frac{1}{2}f'(x) \leq 0$

Question

I am trying to set up a simple (economic) model to illustrate an argument.

I have an increasing concave production function $f(x)=x^\alpha$ with $\alpha\in(0,1)$ and want to show that

$f(x+\frac{1}{2}a) - \frac{1}{2}f(x+a) - \frac{1}{2}f(x) - C \geq 0$

is less likely to be fulfilled as $x$ increases and thought the easiest way would be to show that the first derivative is negative:

$f'(x+\frac{1}{2}a) - \frac{1}{2}f'(x+a) - \frac{1}{2}f'(x) \leq 0$

$\Leftrightarrow (x+\frac{1}{2}a)^{\alpha-1} - \frac{1}{2}(x+a)^{\alpha-1} - \frac{1}{2}(x)^{\alpha-1} \leq 0$

I believe this shouldn't be hard, but I am rusty get stuck, not being able to resolve this further.

Another idea was to show that the limit of the this first derivative is $0$, and that it is strictly increasing in $x$, but I fail to see how the second part could be proven.

I am looking for hints on how to proceed with this, but since I need this mainly to illustrate an argument, I'd also be willing to accept an answer using another $f(x)$ that is strictly increasing and concave, if it significantly simplifies the proof.

Answer 1

$g(x) = f'(x) = \alpha x^{\alpha -1}$ is strictly convex since $$ g''(x) = \alpha (\alpha-1)(\alpha-2) x^{\alpha-3} > 0 $$ for $0 < \alpha < 1$ and $x > 0$. It follows that $$ g(x+ \frac 12 a) < \frac 12 \bigl( g(x) + g(x+a) \bigr) $$ which is the desired inequality.

Answer 2

$$S=f'(x+a/2)-\frac{f'(x)}2-\frac{f'(x+a)}2=g(x+a/2)-g(x)$$ where $$g(x)=\frac{f'(x)-f'(x+a/2)}2.$$ By MVT, $$S=\frac a2g'(y)=\frac a4(f''(y)-f''(y+a/2))$$ where $x<y<a/2$. By MVT again, $$S=-\frac{a^2}8f'''(z)$$ for a suitable $z$. But $f'''(x)=\alpha(\alpha-1)(\alpha-2)x^{\alpha-3}>0$ as $\alpha>0>\alpha-1$.