How to satisfy an inequality, from a fourth degree polynomial

Question

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?

Answer 1

Inspired by https://artofproblemsolving.com/community/c6h590855p3499695 on AoPS:

$$ P(x) = x^4+ax^3+2x^2+bx+1 = x^2 \left(x + \frac a2 \right)^2 + \left(\frac b2 x + 1 \right)^2 + \left( 2 - \frac{a^2+b^2}{4}\right) x^2 $$ has no real zeros if $a^2+b^2 < 8$.

The bound is sharp: For $a=b=2$ we have $a^2+b^2=8$ and $$ P(x) = x^4+2x^3+2x^2+2x+1 = (x+1)^2(x^2+1) $$ with a real zero at $x=-1$.

Answer 2

By Cauchy's inequality, $$(a^{2} + b^{2})(x^{6} + x^{2}) \geq (ax^{3} + bx)^{2} = (-(x^{4}+2x^{2}+1))^{2} = (x^{2}+1)^{4}$$ so $$ a^{2} +b^{2} \geq \frac{(x^{2} + 1)^{4}}{x^{6} + x^{2}} $$ Now try to show that the RHS attains minimum value $8$ when $x = 1$. One way is to use substitution $t=x + \frac{1}{x}$ and then $$ \frac{(x^{2} + 1)^{4}}{x^{6} + x^{2}} = \frac{t^{4}}{t^{2} - 2} \geq 8 \Leftrightarrow (t^{2} - 4)^{2} \geq 0 $$