parmenides51 wrote: Non-zero numbers $a, b, c$ are such that $ax^2+bx+c > cx$ for any $x$. Prove that $cx^2-bx + a > cx-b$ for any $x$. First condition implies $a>0$ and $(b-c)^2-4ac<0$ And so $c>0$ and $(b-c)^2-4ac<0$ may be written $(b+c)^2-4c(a+b)<0$, which is the second condition Q.E.D.