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.
 A: 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
$$ x^4 + ax^3 + 2x^2 + bx + 1>0. $$
A: Multiply though by $4$ to obtain $$4x^4+4ax^2+8x^2+4bx+4=0$$Now note that this can be rewritten $$x^2(2x+a)^2+(bx+2)^2+(8-a^2-b^2)x^2=0$$
If we have $a^2+b^2\lt 8$ then every term is non-negative, and they can't all be zero together (the last term would require $x=0$, but then $(bx+2)^2\gt0$).
A: I assume that $a,b$ ar real.
Suppose that $a^2+b^2<8$ and the polynomial has the real root $\xi$. It follows that:
$$
0=\xi^4 + a\xi^3 + 2\xi^2 + b\xi + 1>\xi^4 + a\xi^3 + \frac{a^2+b^2}{4}\xi^2 + b\xi + 1=\xi^2\left(\xi+\frac{a}{2}\right)^2+\left(\frac{b}{2}\xi+1\right)^2
$$ 
But the sum of squares of real numbers is always positive, therefore $\xi$ has to be complex; contradiction. Thus we have $a^2+b^2\ge8$.
A: Let $x$ be a root.
Thus, by C-S $$(x^2+1)^2=-x(ax^2+b^2)\leq\sqrt{x^2(ax^2+b)^2}\leq\sqrt{x^2(x^4+1)(a^2+b^2)}.$$
Id est, it's enough to prove that:
$$\frac{(x^2+1)^4}{x^2(x^4+1)}\geq8$$ or
$$(x^2-1)^4\geq0.$$
