<s>[INST] Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
 [/INST] Let $f_a(x) = x^4+2ax^3+x^2+2ax+1$. If $x$ is a negative root, then we have 
$$a = -\frac{x^4+x^2+1}{2x^3+2x}$$
That is, if we let 
$$g(x) = \frac{x^4+x^2+1}{2x^3+2x}$$
then $a = g(-x)$. We have that $-x > 0$; using standard techniques, $g(-x)$ reaches a maxiximum (on $(-\infty, 0)$) of $3/4$ at $x=-1$. 
This shows that $a\geq 3/4$. It's easy to show that $f_{3/4}(x)$ only has one negative real root, and so $a > 3/4$. 
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