Assume both of roots of $P(x)$ are nonnegative. Then $P(x)$ can be written into $2$ forms. First form : $P(x)=mx^2-nx+k$ where $m,n,k \in$ $\mathbb{R^+}$ Then $(a^2+b^2)^2m-(a^2+b^2)n+k\geq 4a^2b^2m-2abn+k$ for all $|a|,|b|\geq 2017$ After simplification we get $(a+b)^2m\geq n$ for $a\ne b$. Then $(a+b)^2\geq$ $\frac{n}{m}$ $\geq$ $\frac{4k}{n}$ Since $n$ can't be $0$ get that $(a+b)^2$ is bounded from below different from $0$ and this is contradiction. Second form : $P(x)=-mx^2+nx-k$ where $m,n,k \in$ $\mathbb{R^+}$ Then $-(a^2+b^2)^2m+(a^2+b^2)n-k\geq -4a^2b^2m+2abn-k$ for all $|a|,|b|\geq 2017$ After simplification we get $n\geq (a+b)^2m$ for $a\ne b$ Then $\frac{n}{m}\geq (a+b)^2$ since $\frac{n}{m}$ is constant as $a,b$ goes to infinity we get contradiction again $\blacksquare$

$P(x)=\alpha x^2+\beta x+\gamma $ $P(a^2+b^2) \geq P(2ab)$ so $(a-b)^2(\alpha (a+b)^2+ \beta) \geq 0$ and it is true for $a$ and $b$ such that $|a|,|b| \ge 2017$ Let's selct numbers $a$ and $b$ such that $ a > b$ (we can do it because this inequality is true for every $|a|,|b| \ge 2017$ when $(\alpha (a+b)^2+ \beta) \geq 0$ for instance for $a=2018 , b=-2018$ we get that $\beta \geq 0$ Now suppose that $\alpha < 0$ Then for enough big sum $a+b$ $(a \neq b)$ $(\alpha (a+b)^2+\beta)$ comes to minus infinity(where $\beta > 0$) what is in contradition with fact that $(\alpha (a+b)^2+ \beta) \geq 0$ for any $|a|,|b| \ge 2017$ $a \neq b$ In conclusion we get that $\alpha$ must be positive then $x_1 = \frac{-\beta - \sqrt{\beta^2-4\alpha \gamma}}{2\alpha} < 0 $ where $\sqrt{\beta^2-4\alpha \gamma} > 0$