Solution June 02, 2007

Problem

Find all polynomials of degree 3, such that for each $x,y\geq 0: p(x+y)\geq p(x)+p(y)$

Solution

Let

$p(x):=ax^3+bx^2+cx+d\,$ .

For all $x, y\geq 0$ we have

$a(x+y)^3 + b(x+y)^2 +c(x+y) + d \geq a(x^3+y^3)+b(x^2+y^2)+c(x+y)+2d$

$a(x+y)^3 + b(x+y)^2 \geq a(x+y)(x^2-xy+y^2)+b(x^2+y^2)+d$

$3a(x+y)xy + 2bxy \geq d \qquad \qquad (1)$

First let $x = y = 0$ to see that $d\leq 0$ .

Now set $x = y$ and let $x$ go to infinity we see from (1) that $a\geq 0$ .

Consider the inequality

$q(t) := 6a t^3 + 2bt^2 \geq d \qquad \qquad (2)$

(2) follows from (1) for all $t\geq0$ by letting $x=y=t\,$ . On the other hand, if we let $t=\sqrt{xy}$ then (1) follows from (2) and the arithmetic-geometric mean inequality, $x+y\geq 2t$ .

If $a = 0\,$ , we find that $b\geq0$ by letting $t$ go to infinity, and this suffices to satisfy (2).

Assume that $a>0\,$ . We want to find a condition on $d$ such that (2) is always satisfied. We start by minimizing $q (t)$ for $t\geq 0$ , so we set its derivative to $0$ :