7
$\begingroup$

If it is known that the equation $$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$ has a (real) root, prove the inequality $$ a^2 + b^2 \geq 8. $$

I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway, I can't figure out.

$\endgroup$
11
$\begingroup$

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 $a^2+b^2\ge8$.

$\endgroup$
2
  • $\begingroup$ The sum of squares of real numbers is always non-negative. It remains to show that both squares can't be $0$ at the same time. $\endgroup$ – principal-ideal-domain May 26 '16 at 9:24
  • $\begingroup$ That would still contradict the strict inequality. $\endgroup$ – Joffan May 26 '16 at 9:39
4
$\begingroup$

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 $$ x^4 + ax^3 + 2x^2 + bx + 1>0. $$

$\endgroup$
3
$\begingroup$

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 $(bx+2)^2\gt0$).

$\endgroup$
1
$\begingroup$

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 $$(x^2-1)^4\geq0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.