Proof that inequality involving logarithms and exponentials holds. Let $a\in(0,1)$ and $x\geq 1$. I want to show that
$$g(a,x)=-\ln(a)\cdot a^x\cdot x -1+a^x\leq 0.$$
My attempt: I can define $G(x)=\frac{1}{x}(1-a^x)$ and then write
$$\int_a^1z^{x-1}\,dz=G(x).$$
Then, as $0\leq z^{x-1}\leq 1$ for every $x\geq 1$ and $z\in(0,1)$, the dominated convergence theorem allows me to exchange integral and derivative to get
$$G'(x)=\int_a^1\log(z)z^{x-1}dz\leq 0,$$
Finally, direct computation yields $G'(x)=\frac{g(a,x)}{x^2}$, so we conclude that $g(a,x)\leq 0$.
My question is if anyone have any ideas to show that $g(a,x)\leq 0$ directly. Thanks!
 A: We have $$g(a,x) = -\ln(a)\cdot a^x\cdot x -1+a^x = x\frac{f(x) - f(0)}{x-0} = xf'(\xi)$$
for $f(x) = -\ln(a)\cdot a^x\cdot x +a^x$ and $\xi \in (0,x)$
With $f'(x) = -\ln^2(x)a^x x$ it follows $$g(a,x) = -x\ln^2(\xi)a^\xi \xi$$ for a $\xi \in (0,x)$ hence $g(a,x) \le 0$
A: Here is another direct way just using that $e^t > 1+t$ for $t>0$.
Set $a =\frac 1b$ with $b>1$. Your inequality becomes
$$\ln b\cdot b^{-x}\cdot x - 1 +b^{-x}\leq 0$$
Now, multiply by $b^x$ and rearrange
$$\ln b \stackrel{x>0}{\leq} \frac{b^x-1}{x}$$
which is true because of
$$\frac{b^x-1}{x} = \frac{e^{x\ln b}-1}{x}> \frac{1 + x\ln b - 1}{x} =\ln b$$
Done.
A: Consider the function $g(x) = -\ln(a)a^xx - 1 + a^x$ where $a\in (0,1)$.  Then we have $$g'(x) = -\ln^{2}(a)a^xx - \ln(a)a^x + \ln(a)a^x = -\ln^{2}(a)a^xx.$$  Now, as $a\in (0,1)$, we have that $g'(x) = 0 \implies x = 0$.  Furthermore, $g'(x) > 0$ for all $x\in (-\infty,0)$ and $g'(x)<0$ for all $x\in (0,\infty)$.  Therefore, we have that $x = 0$ is a global maximizer for $g(x)$, and as $g(0) = 0$ we have $g(x) \leq 0$.
