Here's an aggravating problem. Fix some constant $\theta\in(0,1)$. I have a function $$g_\theta(t)=t^\theta\left[(t+1)^{1-\theta}-\left(t+\frac{1}{2}\right)^{1-\theta}\right]$$ which, when plotted in wolfram for various values of $\theta$, is clearly increasing on $[1,\infty)$. But, I need to prove this rigorously.
The obvious thing to do is to try to show that $g'_\theta(t)>0$ on $[1,\infty)$. Unfortunately, we have this mess for the derivative:
$$g'_\theta(t) =(\theta t^{\theta-1}+t^\theta)(t+1)^{-\theta} -(\theta t^{\theta-1}/2+t^\theta)\left(t+\frac{1}{2}\right)^{-\theta}$$
Maybe there's some special function I can use to make this easier. Or some convexity trick I'm not seeing. Idk. What do you guys think?
Thanks!
Applying the mean value theorem to the function $\phi(t,x)=(t+x)^{1-\theta}$ one can show that: $\phi(t,1)-\phi(t,\frac{1}{2})=\frac{1}{2}(1-\theta)(t+\alpha)^{-\theta}$ for some alpha with $ \frac{1}{2}<\alpha<1$. It follows that: $g_{\theta}(t)=\frac{1}{2}(1-\theta)(\frac{t}{t+\alpha})^{\theta}$ which is an increasing function.