A surprisingly resistant elementary numerical inequality Let $a$ and $b$ be real numbers such that $a\geq 1$ and  $b$ lies in the interval $[0,a-1]$. 
How can I then prove that $$(a-b)^{a-b}\geq a^{-b-a}\quad ?$$
This innocent looking inequality seems tougher than I tought: I tried different methods, but none would works: Neither by using calculus to prove that $(a-b)^{a-b} - a^{-b-a}\geq 0$, nor by playing with the inequality and expanding for example the LHS into a binomial series didn't work (or maybe I didn't missed something).
That it holds I'm pretty sure (for said values of $b$): I plotted both side and the inequality holds true for pretty big values of the variables.
 A: I am assuming that $a>2$. Otherwise $[0,\frac{a}{2}-1]$ does not make sense. Since $b\in[0,\frac{a}{2}-1]$, we have $a>2b+2$, which implies that $a-b>b+2$. 
$$ a^{b+a}(a-b)^{a-b}\geq a^{b+a}(b+2)^{b+2}>1 .$$ The last inequality follows from the fact that $b>0$ and $a>2$. Dividing both sides by $a^{a+b}$ we have the result. 
A: We have $0<b<(\frac{a}{2} - 1)$, which we can rewrite as $(a-b) > (b+2) > 2$
Also, $a>=2$ for above inequality to hold.
Since a is clearly positive, let's multiply both sides of the inequality by $a^{a-b}$.
We now just have to prove that $((a-b)*a) ^ {a-b} > 1$
Now, $a-b > 2$ and $a > 2$ . Hence the above product is clearly > 1.
A: From your bounds for $b$ is seems we are assuming $a>2$. If so, it is clear $a>b>0$. Rewriting your inequality we have: $$(a-b)^{a-b}>\frac{1}{a^{b+a}}$$ From your bounds on $b$ we have: $$a-b\geq \frac{a}{2}+1>1$$Furthermore, $b+a$ is clearly greater than $1$. So let's look at what we have: We have a number greater than $1$, that is $a-b$, raised to a power greater than $1$, which is clearly greater than $1$. On the other side we have the reciprocal of a number greater than $1$ raised to a power greater than $1$. This reciprocal is clearly less than $1$. Thus $(a-b)^{a-b}>1$ and $a^{-b-a}<1$ for all $a$, $b$ that fit the conditions and your inequality holds.
