if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four real roots. prove or disprove $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and when $P(x)=x^4+2x^3+2x^2+2x+1=(x+1)^2(x^2+1)$ but if the $P(x)$ have four real roots,I can't it.
Best Answer
Clearly, $x=0$ is not a root. WLOG, assume that $|b| \ge |a|$ (otherwise, letting $x = \frac{1}{y}$, we have $y^4 + by^3 + 2y^2 + ay + 1 = 0$; then swap $a$ and $b$).
Let $x_1 \le x_2 \le x_3 \le x_4$ be the four real roots. We split into two cases.
1) If $x_1x_4 > 0$, then either $x_i>0, \forall i$ or $x_i < 0, \forall i$. By AM-GM, we have $|a| = |x_1+x_2+x_3+x_4| \ge 4\sqrt[4]{x_1x_2x_3x_4} = 4$ and hence $a^2 + b^2 \ge 2a^2 \ge 32$.
2) If $x_1x_4 < 0$, then $x_1 \le x_2 < 0 < x_3 \le x_4$ since $x_1x_2x_3x_4 = 1$. From $x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 + x_1x_3 + x_2x_4 = 2$, we have \begin{align} x_1x_2 + x_3x_4 &= 2 - x_2x_3 - x_4x_1 - x_1x_3 - x_2x_4\\ &\ge 2 + 4\sqrt[4]{(-x_2x_3)\cdot (-x_4x_1) \cdot (-x_1x_3) \cdot (-x_2x_4)}\\ &\ge 6. \end{align} Then, we have \begin{align} a^2 &= (x_1 + x_2 + x_3 + x_4)^2 \\ &= x_1^2 + x_2^2 + x_3^2 + x_4^2 + 2 (x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 + x_1x_3 + x_2x_4)\\ &\ge 2x_1x_2 + 2x_3x_4 + 4\\ &\ge 16. \end{align} Thus, $a^2 + b^2 \ge 2a^2 \ge 32$.
We are done.