Jjesus wrote: Find all functions $f\colon \mathbb{R}\to\mathbb{R}$ such that \[f(x + y) + f(x + z) - f(x)f(y + z) \ge 1\]for all $x,y,z \in \mathbb{R}$ $x=y=z=0$ gives $f(0)=1$. $y=z=0$ gives $f(x) \geq 1$ for all $x$. $y=z=-x$ gives $f(x)f(-2x) \leq 1$ for all $x$. So $f\equiv 1$.