Let $\textbf{Sigmoid}(x) = \sigma(x)=\frac{1}{1+e^{-x}}$. How can we show $\textbf{Sigmoid}(y)$ is concave where $y=\log x + \alpha$ and $\alpha\geq 0$?
My try:
We know $\sigma'(x)=\sigma(x)(1-\sigma(x))=\sigma(x)\sigma(-x)$. Also, we know $$ \textbf{Sigmoid}'(y)=\sigma'(y)=y'\sigma'(y)=\frac{1}{x}\sigma'(y)=\frac{1}{x}\sigma(y)\sigma(-y) $$
Therefore,
\begin{equation} \begin{split} \textbf{Sigmoid}''(y)=\sigma''(y)&=(\frac{1}{x}\sigma(y)\sigma(-y))'\\ &=-\frac{1}{x^2}\sigma(y)\sigma(-y) + \frac{1}{x}y'\sigma'(y)\sigma(-y) +\frac{1}{x}\sigma(y)(-y)'\sigma'(-y)\\ &= -\frac{1}{x^2}\sigma(y)\sigma(-y) + \frac{1}{x^2}\sigma(y)\sigma(-y)\sigma(-y) -\frac{1}{x^2}\sigma(y)\sigma(-y)\sigma(y)\\ &=\frac{1}{x^2}\sigma(y)\sigma(-y) \Big( -1 + \sigma(-y) - \sigma(y) \Big) \end{split} \end{equation}
Since both $\sigma(y)$ and $\sigma(-y)$ are positive and $x>0$, we need to find the sign of the last term.
Please to answer my question in limit sense or intuitively, I want to show the last parentheses is negative.
\begin{equation} \begin{split} -1 + \sigma(-y) - \sigma(y)&=-1+\frac{1}{1+e^{y}} - \frac{1}{1+e^{-y}}\\ &= -1 + \frac{1+e^{-y}}{(1+e^{y})(1+e^{-y})} - \frac{1+e^{y}}{(1+e^{y})(1+e^{-y})}\\ &=-1 + \frac{e^{-y}-e^{y}}{(1+e^{y})(1+e^{-y})} \\ &= \frac{-1 -e^{-y} -e^{y} -1 + e^{-y}-e^{y}}{(1+e^{y})(1+e^{-y})}\\ &=\frac{-2-2e^{y} }{(1+e^{y})(1+e^{-y})} \end{split} \end{equation}