Prove inequalities $\frac {36}{25} \le A(a) < 2$

Question

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.

Answer 1

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: $$\left(a^2+a+\frac{1}{2}\right)(a^2+2a+2)\geq\left(a^2+\sqrt2a+1\right)^2>(a^2+a+1)^2.$$

Answer 2

Note that $$2-A(a)=2\,{\frac {a \left( 2\,{a}^{2}+3\,a+2 \right) }{ \left( 2\,{a}^{2}+2\, a+1 \right) \left( {a}^{2}+2\,a+2 \right) }} >0$$ since $a>0$ and $$A(a)-\frac{36}{25}={\frac { \left( 28\,{a}^{2}+40\,a+28 \right) \left( a-1 \right) ^{2} }{ \left( 50\,{a}^{2}+50\,a+25 \right) \left( {a}^{2}+2\,a+2 \right) }} \geq 0$$ The equal sign holds if $a=1$

Answer 3

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 $a > 0$.

Answer 4

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 $a$.