Prove $a^2+b^2\ge 8$ for real roots of $x^4+ax^3+2x^2+bx+1=0$

[anemone]

Here is this week's POTW:

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Prove that $a^2+b^2\ge 8$ if $x^4+ax^3+2x^2+bx+1=0$ has at least one real root.

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[anemone]

Congratulations to Opalg for his correct solution(Cool), which you can find below:

Spoiler

If $x$ is a real root of $x^4 + ax^3 + 2x^2 + bx + 1 = 0$ then clearly $x$ cannot be $0$. So we can divide by $x^2$ and write the equation as $$\bigl(x+\tfrac1x\bigr)^2 = -\bigl(ax + \tfrac bx\bigr) = \bigl|ax + \tfrac bx\bigr|$$ (because the left side must be positive). By the Cauchy–Schwarz inequality, $$\bigl|ax + \tfrac bx\bigr| \leqslant \sqrt{a^2+b^2}\sqrt{x^2 + \tfrac1{x^2}} = \sqrt{a^2+b^2}\sqrt{\bigl(x+\tfrac1x\bigr)^2 - 2}.$$ Let $y = x+\frac1x$. Then it follows that $y^2 \leqslant \sqrt{a^2+b^2}\sqrt{y^2 - 2}.$

On the other hand, $0\leqslant (y^2-4)^2 = y^4 - 8y^2 + 16$, so that $y^4 \geqslant 8y^2-16$, and therefore $y^2 \geqslant \sqrt{8(y^2-2)}$.

Thus $\sqrt{a^2+b^2}\sqrt{y^2 - 2} \geqslant y^2 \geqslant \sqrt{8(y^2-2)}$. But $y^2 = \bigl(x+\tfrac1x\bigr)^2$, which has minimum value $4$, so that $\sqrt{y^2 - 2}$ is not zero and we can divide by it, to get $\sqrt{a^2+b^2} \geqslant \sqrt8$ and hence $a^2+b^2 \geqslant 8.$

Alternative solution from other:

Spoiler

Multiply though by 4 to obtain $4x^4+4ax^3+8x^2+4bx+4=0$.

Now, note that this can be rewritten
$x^2(2x+a)^2+(bx+2)^2+(8−a^2−b^2)x^2=0$

If we have $a^2+b^2<8$ then every term is non-negative, and they can't all be zero together (the last term would require $x=0$, but then $(bx+2)^2>0$).