<s>[INST] 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.
 [/INST] 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 $a^2+b^2\ge8$.
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