I'm trying to prove that $\frac{1}{(1+e^x)}$ is strictly convex over $[0,\infty)$
Over $(0,\infty)$ it is easily proven using the theorem:
Let $f$ be a twice continuously differentiable function over a convex set $C\subseteq\mathbb{R}^n$ and suppose that $\nabla^2f(x)>0$ for any $x\in C$. Then $f$ is strictly convex of $C$.
However, this theorem doesn't apply to $x=0$ since $f''(x)=\frac{e^x}{(1+e^x)^2 }⋅(\frac{2e^x}{1+e^x}-1)$ and $f''(0)=0$
How then can I show this for $x=0$?
You've already proved that $f$ is strictly convex on $(0,\infty)$. Now all that's left is to prove that for all $x>0$ and $\lambda\in(0,1)$ it is $f(\lambda\cdot x+(1-\lambda)\cdot0)<\lambda f(x)+(1-\lambda)f(0)$, that is $f(\lambda x)<\lambda f(x)+\frac{1-\lambda}{2}$. Let $g(x)=\lambda f(x) +\frac{1-\lambda}{2}-f(\lambda x)$, for $x\geq0$. Then, it is $g(0)=0$ and $g'(x)=\lambda f'(x)-\lambda f'(\lambda x)=\lambda (f'(x)-f'(\lambda x))$. But as you may notice, $\lambda x<x$ and $f'$ is strictly increasing, hence $g'(x)>0$ for $x>0$; therefore $g$ is strictly increasing and $g(x)>0$ for $x>0$.
