How to prove the inequality? Given $0<x<1$, $0<a<b<1$, and $a+b<1$, how to prove $a^x(1-ax)<b^x(1-bx)$?
I've tried using $f(x)=x^t(1-xt)$ to do some manipulations (including derivations), but failed.
 A: The function $f(x)=x^k(1-kx)=x^k-kx^{k+1}$ has derivative
$$f'(x)=kx^{k-1}(1-(k+1)x).$$
So the only critical point of $f$ is at $1/(k+1)$, and the latter is greater than $1/2$.
Now assume $0<x<y$ and $x+y<1$, and form the difference
$$w(x,y)=f(y)-f(x).$$
We wish to show this is nonnegative under the conditions. The inequalities defining the restriction may be made inclusive, and then the region $R$ satisfying  them is bounded by a triangle with vertices $(0,0),\ (1/2,1/2),\ (0,1).$  When the partials of $w$ are each put to $0$ to get a possible critical point, the only one having positive $x,y$ in the containing square $[0,1]^2$ is $(1/(k+1),1/(k+1))$  but this lies outside $R$ since $1/(k+1)>1/2.$ [We do not even have both partials defined on the left vertical side of $R$, since in fact as $x \to 0^+$ we have $f' \to + \infty.$] Therefore the minimum of $w$ must occur on one of the three edges of the triangle $R$.
On the left edge, where $x=0,\ 0 \le y \le 1$ we have $w=f(y) \ge 0.$ On the edge corresponding to $y=x$ we have $w=0.$ The remaining "downwardly slanted" edge may be parametrized by $(x,y)=(t,1-t)$ where $0 \le t \le 1/2.$ We thus turn to considering the function $h(t)=f(1-t)-f(t).$ The direct derivative of $h$ is involved, and I could not get its zeros. However we only need to show that $h'(t)<0$ for $0 \le t \le 1/2,$ for given that, since at $t=1/2$ we have $h(t)=0,$ we could conclude as desired that $h(t)\ge 0$, i.e. that $w \ge 0$ on the last edge.
Now we take the derivative of $h(t)$ (the chain rule introduces two extra minus signs), and rewrite it in a convenient form:
$$\frac{1}{2k}h'(t)=(-1)\cdot \frac{(1-t)^{k-1}+t^{k-1}}{2}+(k+1)\cdot \frac{(1-t)^k+t^k}{2}. \tag{1}$$
We can use the convexity/concavity of the functions $u^{k-1}$ and $u^k$ to obtain an upper bound for $h'(t)/(2k),$ and this upper bound turns out to be negative. First since $u^{k-1}$ is concave up, we have that the first fracion in $(1)$ (without the negative sign in front) is at least $(1/2)^{k-1}.$ Similarly since $u^k$ is concave down the second fraction in $(1)$ is at most $(1/2)^k.$  Putting these together we obtain the upper bound for $h'(t)/(2k)$ of
$$-(1/2)^{k-1}+(k+1)(1/2)^k=(1/2)^{k-1}(-1+(k+1)(1/2))
\\ =(1/2)^{k-1}(k-1)/2<0,$$
the last because of the assumption $0<k<1.$
