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

Solution 1:

I assume that $a,b$ ar real.

Suppose that $a^2+b^2<8$ and the polynomial has the real root $\xi$. It follows that: $$ 0=\xi^4 + a\xi^3 + 2\xi^2 + b\xi + 1>\xi^4 + a\xi^3 + \frac{a^2+b^2}{4}\xi^2 + b\xi + 1=\xi^2\left(\xi+\frac{a}{2}\right)^2+\left(\frac{b}{2}\xi+1\right)^2 $$ But the sum of squares of real numbers is always positive, therefore $\xi$ has to be complex; contradiction. Thus we have $a^2+b^2\ge8$.

Solution 2:

The proof from user109899 is wonderful. Here is another one and the idea is the same. Note \begin{eqnarray} &&x^4 + ax^3 + 2x^2 + bx + 1\\ &=&x^2(x^2+ax)+2x^2+bx+1\\ &=&x^2(x+\frac a2)^2-\frac{a^2}{4}x^2+2x^2+bx+1\\ &=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}x^2+bx+1\\ &=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x^2+\frac{4b}{8-a^2}x\right)+1\\ &=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x+\frac{2b}{8-a^2}\right)^2+1-\frac{b^2}{8-a^2}\\ &=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x+\frac{2b}{8-a^2}\right)^2+\frac{8-a^2-b^2}{8-a^2} \end{eqnarray} Hence if $a^2+b^2<8$, then $8-a^2>0, 8-a^2-b^2>0$ and thus $$ x^4 + ax^3 + 2x^2 + bx + 1>0. $$

Solution 3:

Multiply though by $4$ to obtain $$4x^4+4ax^2+8x^2+4bx+4=0$$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 $a^2+b^2\lt 8$ then every term is non-negative, and they can't all be zero together (the last term would require $x=0$, but then $(bx+2)^2\gt0$).

Solution 4:

Let $x$ be a root.

Thus, by C-S $$(x^2+1)^2=-x(ax^2+b^2)\leq\sqrt{x^2(ax^2+b)^2}\leq\sqrt{x^2(x^4+1)(a^2+b^2)}.$$ Id est, it's enough to prove that: $$\frac{(x^2+1)^4}{x^2(x^4+1)}\geq8$$ or $$(x^2-1)^4\geq0.$$