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

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To prove that \( a^2 + b^2 \ge 8 \) for the polynomial \( x^4 + ax^3 + 2x^2 + bx + 1 = 0 \) having at least one real root, participants discussed various approaches and techniques. The correct solution was provided by Opalg, highlighting the necessary conditions for the coefficients \( a \) and \( b \). Alternative solutions were also shared, showcasing different methods to arrive at the same conclusion. The discussion emphasized the importance of analyzing the polynomial's behavior and its roots. Overall, the problem reinforced key concepts in polynomial inequalities and real root conditions.
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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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Congratulations to Opalg for his correct solution(Cool), which you can find below:
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:
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 $(bx+2)^2>0$).
 
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