Does the inequality $a^{m-1}(m-a)<m-1$, where $0<a<1$, hold for $m=2,3,\cdots$ ?

Question

Does the inequality $a^{m-1}(m-a)<m-1$, where $0<a<1$, hold for $m=2,3,\cdots$ ?

I think it holds (from the numerical results). However, I cannot prove it. Does anyone have some hints?

Answer 1

Check it for $m=2$. You need to prove that

$$a(2-a)<1$$

which is true as it yields to check that $a^2-2a+1 =(a-1)^2>0$.

Now take the derivative of both sides: they respectively are

$$\frac{d}{dm}a^{m-1}(m-a) = a^{m-1}(m-a)\log(a) +a^{m-1}$$

$$\frac{d}{dm} (m-1) = 1$$

Since $\log(a) < 0$ and $m\ge 2$, the derivative of the LHS is lesser than or equal to $a^{m-1}$ which in turn is lesser than $1$ (the RHS derivative, how lucky!). As the derivative of the RHS is always greater than the one of the LHS, then the function on RHS stays always above (beginning from $m=2$ at least) the one on the LHS.

Answer 2

Let $m= 2,3,..$ be fixed.

$f(x)_m := (m-1) - x^{m-1}(m-x),$ $0\le x\le 1.$

We would like to show that for

$m=2,3,..$ and $x \in (0,1)$ :

$ f_m(x) >0.$

At $x=0$, and $x=1:$

$f_m(0) = m-1>0.$

$f_m(1) = 0.$

Now :

$f_m'(x)=$

$-(m-1)x^{m-2}(m-x) +x^m <0.$

Since: $x^{m-2} >x^m$ for $x\in (0,1)$ and $(m-1)(m-x) > 1$, $ m=2,3,..$,

we have :

$(m-1)(m-x) x^{m-2}> x^m$, I.e.

$ f_m'(x) <0 $.

$\rightarrow:$

$ f_m(x)$ is strictly decreasing in $(0,1).$

$\rightarrow:$

$f_m(x) >0$ for $x\in (0,1)$, $m=2,3,..$