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


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