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 ?

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.
