Nartku wrote: Find all second degree polynomials $P(x)$ such that for all $a \in\mathbb{R} , a \geq 1$, then $P(a^2+a) \geq a.P(a+1)$ Let $P(x)=ux^2+vx+w$ and inequation is $u(a^2+a)^2+v(a^2+a)+w\ge a(u(a+1)^2+v(a+1)+w)$ $\forall a\ge 1$ Which is $ua^4+ua^3-ua^2-(u+w)a+w\ge 0$ $\forall a\ge 1$ So $u>0$ and, setting $t=\frac wu$ : $a^4+a^3-a^2-(t+1)a+t\ge 0$ $\forall a\ge 1$ Which is $(a-1)(a^3+2a^2+a-t)\ge 0$ $\forall a\ge 1$ And so $a^3+2a^2+a-t\ge 0$ $\forall a>1$ And since $a^3+2a^2+a-t$ is increasing over $\mathbb R^+$ (at least), this is $a^3+2a^2+a-t\ge 0$ for $a=1$ and so $t\le 4$ And so $\boxed{u>0\text{ and }w\le 4u}$