izat wrote: Find all functions $ f:R \implies R $ , such for all $x,y,z$ $f(xy)+f(xz)\geq f(x)f(yz) + 1$ Let $P(x,y,z)$ be the assertion $f(xy)+f(xz)\ge f(x)f(yz)+1$ $P(0,0,0)$ $\implies$ $f(0)=1$ $P(1,1,1)$ $\implies$ $f(1)=1$ Let $x\ne 0$ : $P(x,0,\frac 1x)$ $\implies$ $1\ge f(x)$ and $P(1,0,x)$ $\implies$ $f(x)\ge 1$ And so $\boxed{f(x)=1\text{ }\forall x}$ which indeed is a solution

Find all functions $f:\mathbb R\to\mathbb R$ such that $$f(xy)+f(xz)-f(x)f(yz)\ge1\qquad\text{for all }x,y,z.$$

$P(0,1,0)\Rightarrow f(0)=1$ $P(1,x,0)\Rightarrow f(x)\ge f(1)$ $P(x,x^{-1},0)\Rightarrow f(x)\le f(1)\Rightarrow f(x)=f(1)$, testing gives $\boxed{f(x)=1}$.