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

If $x^4+ax^3+3x^2+bx+1 \ge 0$ for all real $x$ where $a,b \in R$. Find the maximum value of $(a^2+b^2)$. I tried setting up inequalities in $a$ and $b$ but in the end had a hideous two variable expression whose maxima I had to calculate using Partial Derivatives. There must be a better way of doing it. Thanks!

Solution 1:

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 $max\{a^2+b^2\} = 8(c+2)$