Prove inequalities $\frac {36}{25} \le A(a) < 2$ Given the expression,
$$ A(a) = \frac{\left( 1 +a +\frac 1a\right)^2 }{\left(\frac12+a+a^2\right)\left(\frac12+\frac 1a + \frac{1}{a^2}\right)} $$
with $a > 0$.
Prove the following inequalities:
$$\frac {36}{25}  \le A(a) < 2$$
Note that the bounds are rather tight. I encountered this issue in determining a narrow range of an angle in a geometry problem. I was only able to examine certain limits and identified the correct answer. But, I did not manage to fully prove the above inequalities. 
Since I am not all that versed in dealing with such type of problems and would appreciate if anyone could suggest solutions for the proof.

In the form given, $A(a)$ is pretty hard to reason around. As a first step let's factor out a factor of $\frac{1}{a^2}$ and see if we can coerce the denominator to look more friendly.
$$
A(a) = \frac{\frac{1}{a^2} {(a^2 + a + 1)}^2}{(a^2 + a + 1 - \frac{1}{2})\frac{1}{a^2}(\frac{1}{2}a^2 + a + 1)} = \frac{{(a^2 + a + 1)}^2}{(a^2 + a + 1)(a^2 + a + 1 - \frac{1}{2}a^2) - \frac{1}{2}(\frac{1}{2}a^2 + a + 1)} \\
A(a) = \frac{{(a^2 + a + 1)}^2}{{(a^2 + a + 1)}^2 -\frac{1}{2}\left[a^2(a^2 + a + 1) + (\frac{1}{2}a^2 + a + 1)\right]} = \frac{{(a^2 + a + 1)}^2}{{(a^2 + a + 1)}^2 -\frac{1}{2}(a^4 + a^3 + \frac{3}{2}a^2 + a + 1)}
$$
Now to conclude anything from this let us examine ${(a^2 + a + 1)}^2$, expanded we get $a^4 + 2a^3 + 3a^2 + 2a + 1$, which compared to the negative term in the denominator is almost exactly twice as big. The $-\frac{1}{2}$ enables us to resolutely conclude that our denominator is greater than half the numerator for $a > 0$, so we have $A(a) < 2$. 
Unfortunately for our other bound short of calculating the fairly convoluted derivative of $A(a)$ to conclusively prove $a = 1$ is a critical point, and in fact a minimum, I am not sure of an easier way way. However, proving $a =1$ is a minimum (it is) and plugging it in will give us $A(1) = \frac{36}{25}$, so we have $\frac{36}{25} \leq A(a) < 2$ for $a > 0$.
