$x^4 -ax^3 +2x^2 -bx +1$ has real root $\implies$ $a^2+b^2 \ge 8$

Since the original question has now been resurrected, I may as well post my own thoughts on it. I tried a few things that didn't work, then hit upon the following approach, when $x \ne 0$ let $y=\frac1x$ $$ y^2f(x) = \left(x-\frac{a}2\right)^2 +\left(y-\frac{b}2\right)^2 + 2 - \frac{a^2}4 - \frac{b^2}4 $$

We have from Cauchy-Schwarz inequality: $((1+x^2)^2)^2=(x^4 + 2x^2 + 1)^2 = (ax^3+bx)^2 \le (a^2+b^2)(x^6+x^2) = (a^2+b^2)x^2(x^4+1)\implies a^2+b^2 \ge \dfrac{(1+x^2)^4}{x^2(x^4+1)}\ge 8 \iff (1+x^2)^4 \ge 8x^2(1+x^4) \iff (1+y)^4 \ge 8y(1+y^2), y = x^2 \ge 0$. Lastly, consider $f(y) = (1+y)^4 - 8y(1+y^2), y \ge 0\implies f'(y) = 4(1+y)^3 - 8-24y^2 = 4((1+y)^3 - 2 - 6y^2)= 4(y-1)^3$. Thus if $0 \le y \le 1 \implies f'(y) \le 0 \implies f(y) \ge f(1) = 0$. If $y \ge 1 \implies f'(y) \ge 0 \implies f(y) \ge f(1) = 0$. Either case $f(y) \ge 0\implies a^2+b^2 \ge 8$ as claimed.

Let $x$ be a root. Thus, $x\neq0$ and $b=\frac{x^4-ax^3+2x^2+1}{x}$ and we need to prove that $$a^2+\frac{(x^4-ax^3+2x^2+1)^2}{x^2}\geq8$$ or $$(x^6+x^2)a^2-2(x^7+2x^5+x^3)a+x^8+4x^6+6x^4-4x^2+1\geq0,$$ for which it's enough to prove that $$(x^7+2x^5+x^3)^2-(x^6+x^2)(x^8+4x^6+6x^4-4x^2+1)\leq0$$ or $$(x^2-1)^4\geq0.$$ Done!