Show a function is bounded below by $-\frac{b^2}{4a}$

Question

The question asks to show that the infimum of the function $p(x) = 2x^4 + ax^2 +bx$ is always above the expression $-\frac{b^2}{4a}$ where $b$ is any number, and $a$ is nonnegative.

To make sense, I guess we cannot allow $a$ to be zero for the above expression to make sense.

My attempt:

Since the function is to be bounded below, I claimed that $$-\frac{b^2}{4a} \le 2x^4 + ax^2 +bx$$ and then I move things around to get:

$$0 \le 8ax^4 + 4a^2x^2 + 4abx \; +b^2$$

Now, I saw that $8ax^4 + 4a^2x^2 + b^2 $ must always be nonnegative, so I would just have to show that $|4abx| \le 8ax^4 + 4a^2x^2 +b^2$ and then the above inequality is sure to always hold... However, I'm kind of stuck. I know that my claim is true because I graphed it and it works, but I keep running around in circles trying to actually show it rigorously.

Or maybe I'm not even thinking about it correctly at all...

Thank you for your input!!!

Answer 1

You have to prove $$0 \le 8ax^4 + 4a^2x^2 + 4abx \; +b^2$$ From your last line. Using the SOS method, note that $$8ax^4 + 4a^2x^2 + 4abx \; +b^2=8ax^4+(2ax+b)^2 \ge 0$$

Now since $$(2ax+b)^2 \ge 0$$ and $$8ax^4 \ge 0$$ We are done.