TITLE: Prove that $\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}$ is a strictly decreasing function.
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: Differentiating is a clumsy way of solving the problem. However, let's look at the derivative. It is equal to 
$$-\frac{x(0.5x+1)}{(x^2+x+1)^2}.$$
The denominator is bounded away from $0$. The numerator is negative for $x\gt 0$. Thus (Mean Value Theorem) our function is strictly decreasing in the interval $(0,\infty)$, indeed in the interval $[0,\infty)$.