<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 $x < 0$ be a real root of $x^4+2ax^3+x^2+2ax+1=0$. Since there are two different
negative real roots we can assume that $x \ne -1$.
You already found out that $t=x+ \frac{1}{x}$ satisfies $t^2+2at-1=0$. Then
$$
 -t = \lvert x \rvert + \frac{1}{\lvert x \rvert} > 2
$$
from the AM-GM inequality, with strict inequality since $\lvert x \rvert \ne 1$, therefore $t < -2$.
It follows that 
$$
 a = \frac{1-t^2}{2t} = \frac 12 \left( \frac 1t - t \right) 
 > \frac 34
$$
because the expression is strictly decreasing for $t < 0$.
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