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.

  • $\begingroup$ Very clean proof. :) Thank you. $\endgroup$ – Clarinetist Jun 28 '14 at 2:36

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 $[0,\infty)$.


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