TITLE: How to find the max value of $a^2+b^2$ if $x^4+ax^3+3x^2+bx+1\ge0$
QUESTION [9 upvotes]: Let $f(x)=x^4+ax^3+3x^2+bx+1$, where $a,b \in \mathbb{R}$. If $f(x) \ge 0$ for all $x \in \mathbb{R}$, what is the maximum possible value of $a^2+b^2$?
I don't know how to proceed. Hints or help will be appreciated.

REPLY [2 votes]: Following mathlove's strategy, we can do a little better . . .

If $f(x) = x^4 + ax^3 + 3x^2 + bx + 1$, where
\begin{align*}
a&=0\\[8pt]
b&=\sqrt{6+{\small{\frac{14}{9}}}\sqrt{21}}
\end{align*}
then $f(x) \ge 0$ for all $x \in \mathbb{R}$, and
$$a^2 + b^2= 6+{\small{\frac{14}{9}}}\sqrt{21} \approx 13.12845108$$
I don't know if this is best possible, however it is best possible for the case $a=0$.