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 ?

If we solve $\quad x^4+2ax^3+x^2+2ax+1=0\quad$ for  $a$ we get
\begin{equation}
a = -\frac{x^4 + x^2 + 1}{2 x^3 + x}\quad \text{for}\quad  x^3 + x\ne0
\end{equation}
which suggests that $\quad a<0\quad $ but
$\quad x=1\implies a=-\frac{3}{4}\quad $ and $\quad x=-1\implies a=\frac{3}{4}$
If we simply try a range of values, for $F(x)$ above, we get
\begin{equation}
F(-5)=2.50384615384615\qquad 
F(-4)=2.00735294117647\qquad 
F(-3)=1.51666666666667\qquad 
F(-2)=1.05\qquad 
F(-1)=0.75\qquad 
F(0)=???\qquad 
F(1)=-0.75\qquad 
F(2)=-1.05\qquad \\
F(3)=-1.51666666666667\qquad 
F(4)=-2.00735294117647\qquad 
F(5)=-2.50384615384615\qquad \end{equation}
fra
And this shows that  $\qquad a>\frac{3}{4}\iff x\le -1$
