Upper bound of a function 2 Consider function
$f:\mathbb{R}_{\geq 0} \times [0,1] \rightarrow \mathbb{R}$ defined with:
$$ f(z,\alpha) = (1+\alpha) \left [ z- \frac{z^{\alpha}}{2} - z^{\alpha+1} \right].$$
Prove that $f(z,\alpha) <\frac{1}{2}$ on its domain ($z \geq 0, \alpha \in [0,1]$).  Numerically, the claim  seems to be correct. I'm unable to prove it analytically. 
 A: Here's an analytic solution that I hope is not too convoluted:
First, observe that when $z > 1$, we have:
\begin{align} 
f(z,a) &= (1 + \alpha) (z - \frac{z^{\alpha}}{2} - z^{1+\alpha}) \\
&<(1 + \alpha) \Big(- \frac{z^{\alpha}}{2}\Big) \\
&< 0
\end{align}
So we only need to be concerned when $z \leq 1$. When $z \leq 1$, we have:
\begin{align}
f(z,\alpha) &= (1 + \alpha) (z - \frac{z^{\alpha}}{2} - z^{1+\alpha}) \\
&\leq (1 + \alpha) (z - z^{1+\alpha}) \\
&:= g(z, \alpha)
\end{align}
Now, differentiating $g$ with respect to $z$ and setting to zero (essentially finding the maximum) gives us the following value of $z$:
$$z = (\alpha+1)^{-\frac{1}{\alpha}} $$
So, substituting this back into $g(z,\alpha)$, we obtain:
\begin{align}
g(z, \alpha) &\leq (1+\alpha) ((\alpha+1)^{-\frac{1}{\alpha}} - (\alpha+1)^{-\frac{1}{\alpha}(\alpha + 1)} ) \\
&= \frac{\alpha}{ (1 + \alpha)^{\frac{1}{\alpha} } }\\
&\leq \frac{\alpha}{1 + \alpha} \\
&\leq 0.5
\end{align}
Where the last inequality follows from the fact that $\frac{\alpha}{1+\alpha}$ is an increasing function and is equal to $0.5$ when $\alpha = 1$. 
