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.

Here's a simple approach: let $b = 1 + a + 1/a$, so $b \geq 3$, and
$$\frac{1}{A(a)} = \frac{(ab - 1/2)(b/a - 1/2)}{b^2} = 1 - \frac{a + 1/a}{2b} + \frac{1}{4b^2} = 1 - \frac{b-1}{2b} + \frac{1}{4b^2} = \frac{1}{2} + \frac{1}{2b} + \frac{1}{4b^2}.$$
As $a$ varies over $(0, \infty)$, $b$ varies over $[3, \infty)$, and it's clear that $1/A(a)$ is strictly decreasing as a function of $b$, so since $1/A(a) = 25/36$ at $b = 3$, and $1/A(a)$ decreases to $1/2$ as $b \to \infty$, we have $1/2 < 1/A(a) \leq 25/36$ for all $a$.
