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!!!
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.