Prove $b > 0 \implies b^3 + 3b > 3b^2$ in any ordered field

Question

Prove $b > 0 \implies b^3 + 3b > 3b^2$ in any ordered field. Prove this algebraically (i.e. without continuity or IVT).

Using analysis, this is easy. Is there a way to prove it without analysis or limits, but simply using algebra?

I believe there must be, because I believe that any ordered field would have this property, even if it is incomplete.

If not: does this imply that there exists ordered fields which do not have this property?

Answer 1

You have $$ b^2-3b+3=b^2-3b+\frac94+3-\frac94 =(b-\frac32)^2+\frac34>0. $$ Since $b>0$ multiplying by $b$ preserves the inequality. Then $$ b^3-3b^2+3b>0. $$

Answer 2

Alternatively we want to show that $b^3 > 3b^2-3b = 3b(b-1)$. We can eliminate $b < 1$ easily since $b^3 > 0$ but if $b -1 < 0$ we have $3b(b-1) < 0$ and it is also trivial to check when $b=1$. This leaves the case when $b > 1$ and so we write $b=c+1$ for some $c > 0$. Now substituting gives $$3b^2-3b=3(c^2+2c+1)-3(c+1) = 3c^2-c < c^3 + 3c^2 + 3c+ 1 = (c+1)^3=b^3$$

And so we have $3b^2 -3b < b^3$ as desired.

Answer 3

Another way to approach it.

Since $b > 0$, we can apply the AM-GM inequality to conclude the desired claim: \begin{align*} b + \frac{3}{b} \geq 2\sqrt{3} > 3 \Rightarrow b + \frac{3}{b} > 3 & \Rightarrow b^{3} + 3b > 3b^{2} \end{align*} and we are done.

Hopefully this help!