Solution June 02, 2007


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




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:

18at^2 + 4bt = 0\,
t_1 = 0,\, t_2 = -\frac{2b}{9a}

Note that q(0)=0\geq d so only minima greater than 0 need to be considered.

If b\geq0, then t_2\leq0, so there are no other minima to consider.

If b<0\,, then

q(t_2) = -\frac{16b^3}{243a^2} + \frac{8b^3}{81a^3} = \frac{8b^3}{243a^2} \geq d

\frac{8b^3}{243a^2} is negative in this case, so this is a stronger condition on d than we already had.

In summary, p(x) is a polynomial satisfying p(x+y)\geq p(x)+p(y) for all x,y\geq 0 in the following cases:

No other possibilities for p(x) exist.