Proof or disproof an inequality problem Known:
$$0<x<1$$
$$0<k<1$$
$$0<x+k<1$$
$$y\geq1$$ $y$ is a whole number.
Prove:
$$
\frac{xy}{1-(1-x)^y}\leq \frac{(x+k)y}{1-(1-(x+k))^y}.
$$
I ran some sample results and the inequality appears to be true. For example:
$((0.005*20)/(1-(1-0.005)^9))<((0.006*20)/(1-(1-0.006)^9 ))
$
Basically, both denominators and numerators increase slightly from left to right, which is not obvious whether or not this will increase the right side. It is obvious that denominators on both sides will be between 0 and 1.
Since both denominators have to be positive, I first thought about cross-multiplying but could not countinue to simply further:
 $$
(1-(1-x)^y) ((x+k)y)\leq (xy) (1-(1-(x+k))^y)
$$
 A: Hint: We can reframe the problem to showing that the function
$$f_a(x):=\frac{ax}{1-(1-x)^a}$$
is either monotonely increasing or constant on $[0,1]$, which is the same as showing that $f'_a(x)\geq0$. Differentiating:
$$\begin{split}
f'_a(x) = \frac{a(1-(1-x)^a)-a^2x(1-x)^{a-1}}{(1-(1-x)^a)^2}.
\end{split}$$
The denominator is always positive, so we are reduced to showing that whenever $x\in(0,1)$ and $a\geq1$, then
$$ax(1-x)^{a-1}\leq 1-(1-x)^a.$$
A: $$\frac{xy}{1-(1-x)^y}\leq \frac{(x+k)y}{1-(1-(x+k))^y}$$
Introduce $z=x+k$, $z\in(0,1)$. The inequality becomes:
$$\frac{x}{1-(1-x)^y}\leq \frac{z}{1-(1-z)^y}$$
Make another substitution:
$$u=1-x\\v=1-z$$
Obviously, $x<z\implies u>v$. The inequality now becomes:
$${1-u \over 1-u^y}\le{1-v \over 1-v^y}$$
$${1-u \over {(1-u)(1+u+u^2+...+u^{y-1})}}\le {1-v \over {(1-v)(1+v+v^2+...+v^{y-1})}}$$
$${1 \over {1+u+u^2+...+u^{y-1}}}\le {1 \over {1+v+v^2+...+v^{y-1}}}$$
$$1+u+u^2+...+u^{y-1}\ge 1+v+v^2+...+v^{y-1}$$
...which is obviously true for $u>v$ and $y\ge1$.
