If the equation $P(x)=x^4+ax^3+2x^2+bx+1=0$ has real solutions,
prove that $a^2+b^2\geq8$.
For the above question, I tried taking advantage of Vieta's formulas such as for the roots of the equation $r_1, r_2 , r_3 , r_4$, $r_1+r_2+r_3+r_4=-a/1$ and $r_1r_2+r_1r_3+r_1r_4+...r_3r_4=2/1$ etc. and afterwards I attempted to formulate a quadratic equation. Through this I attempted to get an inequality of the type $b^2-4ac\geq0$, so that I could get that $a^2+b^2\geq8$. However I did not succeed in doing this and hence solving the question. Can someone please explain to me how this question could be solved and if there is a better approach than the one I attempted to use?