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 ?
 A: 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 $t < 0$.
A: 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 $a > 3/4$. 
A: Another approach to determine the behaviour of $f(x)=x^4 + 2ax^3 + x^2 + 2ax + 1$ is by quartic equation analysis.
Define $a,b,c,d,e$ as the coefficients of $f(x)$. 
$$a=1,b=2k,c=1,d=2k,e=1$$
where you want to prove that $f(x)$ has two real negative roots when $k>\frac{3}{4}$ 
Define the discriminant which will allow us to determine the nature of the roots by analyzing its sign: $$\Delta_{f_a} = 256 a^3e^3 - 192 a^2bde^2 - 128a^2c^2e^2 + 144a^2cd^2e - 27a^2d^4 + 144ab^2ce^2 - 6ab^2d^2e - 80abc^2de + 18 abcd^3 + 16ac^4e - 4ac^3d^2 - 27b^4 e^2 + 18b^3cde - 4b^3d^3 - 4b^2c^3e + b^2 c^2 d^2$$
Define $P,Q,D$ which will be useful to evaluate if $f(x)$ will have 4 complex-roots or two negative real roots + two complex conjugate roots.
$$P=8ac - 3 b ^ 2$$ $$Q= b^3 + 8da^2 - 4abc$$ $$D=64 a^3e - 16a^2c^2 + 16 ab^2c - 16a^2bd - 3 b^4$$
Now let's test when $k<\frac{3}{4}$ and $k>\frac{3}{4}$
$$k=0.74<\frac{3}{4}$$
$$P=1.4288, Q=9.16179, D=33.6064, \Delta_{f_{0.74}}=9.13574$$
Since $\Delta_{f_{0.74}}>0, P>0, D>0$ there are two pair of complex roots.
$$k=0.76>\frac{3}{4}$$
$$\Delta_{f_{0.76}}=-9.62079$$
Since $\Delta_{f_{0.76}}<0$ $f(x)$ has two distinct real roots and two complex conjugate root. By Descarte's rule of signs we determine that $f(x)$ has two real negative roots.
Thus when $k>\frac{3}{4}$ $f(x)$ will have two real negative roots.
A: 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  $\qquad a>\frac{3}{4}\iff x\le -1$
