If $x^4 + ax^3 + 3x^2 +bx +1\geq0$, $\forall x\in \mathbb{R}$, find maximum of $a^2+b^2$

Question

Given that $x^4 + ax^3 + 3x^2 +bx +1$ is always greater than equal to $0$ for all $x$ belongs to $\mathbb R$, find $\max(a^2 + b^2)$.

What I did was to show that that above expression is equivalent to $$[x(x+a/2)]^2 + [(12-a^2)(x + 2b/(12-a^2))^2]/4 + (12-a^2 - b^2) /(12-a^2)\geq 0,$$ from this one case is that if $ 12-a^2\geq 0$, then the expression trivially is greater/equal zero, which given max of $a^2+b^2$ to be $12$, but what about if $12-a^2<=0$, then is max of $a^2+ b^2$ greater than 12 possible and what is the exact maximum ? Conclusion: there are two such $f(x)$ for which maxima is attained those are $f(x) = x^4 -2√5x^3 + 3x^2+2√5x +1$ and other as said by achille hui.

Answer 1

I have started using MathJax but am not fluent in using it so posting the solution like this.

Answer 2

Remark: According to @achille hui's comment $x^4 + 2\sqrt{5}x^3 + 3x^2 - 2\sqrt{5}x + 1 = (x^2 + \sqrt{5}\, x - 1)^2$, we give the following solution:

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Let $f(x) = x^4 + ax^3 + 3x^2 + bx + 1$.

Fact 1: If $f(x)\ge 0$ for all $x\in \mathbb{R}$, then $a^2 + b^2 \le 40$.
(The proof is given at the end.)

By Fact 1, we have $a^2 + b^2 \le 40$. On the other hand, if $a = 2\sqrt{5}, ~ b = -2\sqrt5$ with $a^2 + b^2 = 40$, we have $x^4 + 2\sqrt{5}\, x^3 + 3x^2 - 2\sqrt{5}\, x + 1 = (x^2 + \sqrt{5}x - 1)^2 \ge 0$ for all $x\in \mathbb{R}$.

Thus, the maximum of $a^2 + b^2$ is $40$.

We are done.

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Proof of Fact 1: We split into two cases:

Case 1: If $ab \ge 0$, since $f(1) = a + b + 5 \ge 0$ and $f(-1) = - (a + b) + 5 \ge 0$, we have $-5 \le a + b \le 5$ and thus $a^2 + b^2 \le (a + b)^2 \le 25 < 40$.

Case 2: If $ab < 0$, WLOG, assume that $a > 0, ~ b < 0$ (otherwise, $a \to -a, ~ b \to -b, ~ x \to - x$).

Let $x_1 = \frac{-\sqrt5 + 3}{2} > 0$ and $x_2 = \frac{-\sqrt5 - 3}{2} < 0$.
(Note: $x_1$ and $x_2$ are the roots of $x^2 + \sqrt{5}\, x - 1 = 0$. See @achille hui's comment.)

From $f(x_1)\ge 0$, we have $$b \ge -x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1}. \tag{1}$$

From $f(x_2)\ge 0$, we have $$b \le -x_2^3 - ax_2^2 - 3x_2 - \frac{1}{x_2}. \tag{2}$$

From (1) and (2), we have $$-x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1} \le -x_2^3 - ax_2^2 - 3x_2 - \frac{1}{x_2}$$ which results in (note: $x_1 + x_2 = -\sqrt5, x_1x_2 = -1, x_1 - x_2 = 3$) $$a \le \frac{x_1^3 - x_2^3 + 3x_1 - 3x_2 + \frac{1}{x_1} - \frac{1}{x_2}}{x_2^2 - x_1^2} = \frac{x_1^2 + x_1x_2 + x_2^2 + 3 - \frac{1}{x_1x_2}}{-(x_1 + x_2)} = 2\sqrt{5}.$$

Then, using $-x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1} \le b < 0$ and $0 < a \le 2\sqrt5$, we have \begin{align*} a^2 + b^2 &\le a^2 + \left(-x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1}\right)^2\\ &\le (2\sqrt{5})^2 + \left(x_1^3 + 2\sqrt{5}\cdot x_1^2 + 3x_1 + \frac{1}{x_1}\right)^2\\ &= 40. \end{align*} Note: Here, we may use $x_1^2 + \sqrt{5}\, x_1 - 1 = 0$ to reduce the calculations. For example, $x_1^3 = -\sqrt{5}\, x_1^2 + x_1$.

We are done.