Polynomial maximization: If $x^4+ax^3+3x^2+bx+1 \ge 0$, find the maximum value of $a^2+b^2$

I've spent probably 4-5 hours studying this problem and this is what I found:

$$x^4 \pm 2\sqrt{c+2}x^3 + cx^2 \mp 2\sqrt{c+2}x + 1 \ge 0; \forall x,c \in \mathbb{R}$$

Which is true due to the fact that:

$$x^4 - 2\sqrt{c+2}x^3 + cx^2 + 2\sqrt{c+2}x + 1= (-x^2 + \sqrt{c+2}x + 1)^2 \ge 0$$

$$x^4 + 2\sqrt{c+2}x^3 + cx^2 - 2\sqrt{c+2}x + 1= (x^2 + \sqrt{c+2}x - 1)^2 \ge 0$$

So in our case $c=3$ so we have:

$$x^4 \pm 2\sqrt{5}x^3 + 3x^2 \mp2\sqrt{5}x + 1 \ge 0; \forall x \in \mathbb{R}$$

And it's fairly easy to check that the coefficient infront of $x^3$ and $x$ can't have a bigger absolute value. Assume otherwise and we have:

$$x^4 + (2\sqrt{5} + m)x^3 + 3x^2 - (2\sqrt{5}+n)x + 1= (x^2 + \sqrt{5}x - 1)^2 +mx^3-nx$$

Now plug $x_1=\frac{3 - \sqrt{5}}{2}$ and we must have:

$$mx_1^3 - nx_1 \ge 0$$ $$mx_1^3 \ge nx_1$$ $$mx_1^2 \ge n$$ $$m\cdot \frac{7-3\sqrt{5}}{2} \ge n$$

Now plug $x_2=\frac{-3 - \sqrt{5}}{2}$ and we must have:

$$mx_2^3 - nx_2 \ge 0$$ $$mx_2^3 \ge nx_2$$ $$mx_2^2 \le n$$ $$m\cdot \frac{7+3\sqrt{5}}{2} \le n$$

Which clearly contradicts the previous statement. We get simular results if we try to make one of the coefficient bigger, while we are making the other smaller.

From all these contradictions we get $\mid a,b \mid \le 2\sqrt{5}$, so therefore:

$$a^2 + b^2 \le (2\sqrt{5})^2 + (2\sqrt{5})^2 = 20 + 20 = 40$$

It can be simularly proven that if: $x^4 + ax^3 + cx^2 + bx + 1 \ge 0; \forall x \in \mathbb{R}$ for fixed c, then $max\{a^2+b^2\} = 8(c+2)$