Prove that $a^2 + b^2 \geq 8$ if $ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $ has at least one real root. If it is known that the equation
$$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$
has a (real) root, prove the inequality
$$ a^2 + b^2 \geq 8. $$
I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway, I can't figure out.

The proof from user109899 is wonderful. Here is another one and the idea is the same.
Note
\begin{eqnarray}
&&x^4 + ax^3 + 2x^2 + bx + 1\\
&=&x^2(x^2+ax)+2x^2+bx+1\\
&=&x^2(x+\frac a2)^2-\frac{a^2}{4}x^2+2x^2+bx+1\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}x^2+bx+1\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x^2+\frac{4b}{8-a^2}x\right)+1\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x+\frac{2b}{8-a^2}\right)^2+1-\frac{b^2}{8-a^2}\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x+\frac{2b}{8-a^2}\right)^2+\frac{8-a^2-b^2}{8-a^2}
\end{eqnarray}
Hence if $a^2+b^2<8$, then $8-a^2>0, 8-a^2-b^2>0$ and thus
$$ x^4 + ax^3 + 2x^2 + bx + 1>0. $$
