# 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!

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

I’m still wondering about the “beautiful geometric solution,” but there is another intuition that I think is useful.
The idea is that if $f(x) = x^4 + ax^3 + 3x^2 + bx + 1 \geq 0,$
then $f(x)$ can have only multiple roots (else its graph would cross the $x$ axis). A root of multiplicity $3$ is impossible (since it would imply another single root), and it is easy to show that a root of multiplicity $4$ also is impossible (since the terms $x^4$ and $1$ imply the root would have to occur at $x=\pm1,$ and this is inconsistent with the therm $3x^2$). Hence we have two or fewer roots of multiplicity $2.$

Briefly considering how $f(x)$ changes as we vary $a$ and $b,$
if $f(x) \geq 0$ for some pair of values of $a$ and $b$ of the same sign,
then we can reverse the sign of one of those coefficients while preserving
the property that $f(x) \geq 0.$ Hence whatever the maximum value of
$a^2 + b^2$ is, $a$ and $b$ have opposite signs.
The maximum value occurs when there is a double root on each side of the $y$ axis, since roots on only one side would enable one of the coefficients (either the positive one or the negative one) to have a larger magnitude.

In short, $f(x) = k(x – p)^2(x – q)^2$ for some positive number $k$ and some real numbers $p$ and $q.$
Multiplying this out, we have two ways to write $f(x)$ as a polynomial
in standard form,
\begin{align}
f(x) &= kx^4 – 2k(p + q)x^3 + (p^2 + 4pq + q^2)x^2 – 2kpq(p + q)x + kp^2q^2\\
&= x^4 + ax^3 + 3x^2 + bx + 1.
\end{align}

Comparing leading terms, we find that $k=1,$ and therefore the constant terms show that $p^2q^2 = 1,$ that is, $\lvert pq \rvert = 1$.
From the terms in $x$ and $x^3$ we see that $b = pqa$;
since $a$ and $b$ have opposite signs, $pq = -1$ and $q = -\frac1p.$
We can therefore equate the terms in $x^2$ as follows:
$$p^2 – 4 + \left(\frac1p\right)^2 = 3.$$
Setting $u=p^2,$ rearranging terms and multiplying by $u,$
we have $u^2 – 7u + 1 = 0,$ which has roots
$u = \frac12(7 \pm 3\sqrt5).$ We may recognize $\frac12(7 + 3\sqrt5)$ as the fourth power of the golden ratio.
Since $p = \pm\sqrt u,$ $p = \pm\frac12(3 \pm \sqrt5).$
Therefore $p$ and $q$ are either
$\frac12(3 + \sqrt5)$ and $-\frac12(3 – \sqrt5)$
or $-\frac12(3 + \sqrt5)$ and $\frac12(3 – \sqrt5)$ (in either order);
so $p + q = \pm\sqrt5,$ $a = \pm2\sqrt5,$ and $b = -a.$
Therefore $a^2 + b^2 = 40.$