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.

The left inequality.
We need to prove that
$$25(a^2+a+1)^2\geq9(2a^2+2a+1)(a^2+2a+2)$$ and since by AM-GM
$$(2a^2+2a+1)(a^2+2a+2)\leq\left(\frac{2a^2+2a+1+a^2+2a+2}{2}\right)^2,$$ it's enough to prove that
$$100(a^2+a+1)^2\geq9(3a^2+4a+3)^2$$ or
$$10(a^2+a+1)\geq3(3a^2+4a+3)$$ or
$$(a-1)^2\geq0$$ and we are done.
The right inequality.
We need to prove that
$$\left(a^2+a+\frac{1}{2}\right)(a^2+2a+2)>(a^2+a+1)^2,$$ which is true by C-S:
$$\left(a^2+a+\frac{1}{2}\right)(a^2+2a+2)\geq\left(a^2+\sqrt2a+1\right)^2>(a^2+a+1)^2.$$
