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!

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 $a^2 + b^2 = 40.$
