Prove that $\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}$ is a strictly decreasing function. This is part of an actuarial science problem. Unfortunately, the official solution of this problem takes the derivative of 
$$\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}\text{, } \quad x \geq 0\text{.}$$
and shows that it is always $\leq 0$. However, this does not at all show that the function is strictly decreasing.
I'm trying to prove this myself. If I assume $x > y$, I want to show that 
$$\dfrac{0.5x^2 + x + 1}{x^2 + x + 1} < \dfrac{0.5y^2 + y + 1}{y^2 + y + 1}\text{.}$$
Needless to say, this does not look clean if I were to "work backwards."
Any suggestions?

We can write the function as $\dfrac{0.5x^2+x+1}{x^2+x+1} = 1 - \dfrac{0.5x^2}{x^2+x+1} = 1 - \dfrac{0.5}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}$. 
You can easily see that $1+\dfrac{1}{x}+\dfrac{1}{x^2}$ is strictly decreasing, so $\dfrac{0.5}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}$ is strictly increasing. 
Therefore,  $1 - \dfrac{0.5}{1+\dfrac{1}{x}+\dfrac{1}{x^2}} = \dfrac{0.5x^2+x+1}{x^2+x+1}$ is strictly decreasing. 
