Showing that $f(x) \leq 0$ on an interval Let $m \in (0,1)$ and $k \in \mathbb{N}$, and consider
$$f(x) = \frac{1}{k^{m+1}} \frac{1-m}{2(m+1)} - \frac{1-m}{m}\frac{1}{k^m}x.$$
Is there any easy way to show that $f(x) \leq 0$ when $x \in (\frac{1}{k}, k)$?
I can just about do it by graphing but it is painful!!
 A: Since $x>\frac{1}{k}$ you have
$$f(x) < \frac{1}{k^{m+1}} \frac{1-m}{2m+2} - \frac{1-m}{m} \frac{1}{k^m} \frac{1}{k} \\
= \frac{\frac{1-m}{2m+2} - \frac{1-m}{m}}{k^{m+1}}$$
because $f$ is decreasing (it has a negative slope).
Now you just need to show that the numerator is negative. You can prove this formally by getting a common denominator in the big numerator. Intuitively it should be true because you have the same (positive) numerators in the big numerator, so the first fraction is smaller than the second because it has the larger denominator.
A: You have $f(x)=ax+b$.
If $a<0$ then $f(x)\le0$ whenever $x\ge\dfrac{-b}a$.
If $a>0$ then $f(x)\le0$ whenever $x\le\dfrac{-b}a$.
A: This is just algebra really. Let's just make the expression more palatable.
$$f(x) = \frac{1}{2}\frac{1}{k^{m+1}}\frac{1-m}{m+1} - \frac{1}{k^m}\frac{1-m}{m}x$$
Since $k > 0$ we may multiply by $k^m$ without affecting the sign.
$$\frac{1}{2k}\frac{1-m}{m+1} - \frac{1-m}{m}x$$
And $1-m > 0$ so similarly you can cancel this without affecting the sign.
$$\frac{1}{2k}\frac{1}{m+1} - \frac{1}{m}x$$
Can you show that expression is nonpositive? I would multiply out the denominator (which is positive) next.
A: You can simplify:
$$f(x) = \frac{1-m}{k^m}\left(\frac{1}{2k(m+1)} - \frac{x}{m}\right)$$
or
$$f(x) = \frac{1-m}{k^m}\frac{m - 2k(m+1)x}{2km(m+1)}.$$
The first term is always positive, and so is the denominator in the second term. So this is negative exactly when
$$2k(m+1)x > m$$
or
$$x > \frac{1}{2k}\frac{m}{m+1}.$$
Since $\frac 1 k > \frac 1 2 \frac{m}{m+1} \frac 1 k$,
this always holds in the given interval $\left(\frac 1 k, k\right)$.
