A non-increasing particular sequence For every fixed $t\ge 0$ I need to prove that the sequence $\big\{n\big(t^{\frac{1}{n}}-1\big) \big\}_{n\in \Bbb N}$ is non-increasing, i.e.
$$n\big(t^{\frac{1}{n}}-1\big)\ge (n+1)\big(t^{\frac{1}{n+1}}-1\big)\;\ \forall n\in \Bbb N$$
I'm trying by induction over $n$, but got stuck in the proof for $n+1$:

For n=2 its clear that follows since
$$t-1\ge 2(t^{1/2}-1)\Leftrightarrow t-1\ge 2t^{1/2}-2\Leftrightarrow t+1\ge 2t^{1/2}\Leftrightarrow t^2+2t+1\ge 4t\Leftrightarrow t^2-2t+1\ge 0\Leftrightarrow (t-1)^2\ge 0$$
So, we suppose that $\;\ n\big(t^{\frac{1}{n}}-1\big)\ge (n+1)\big(t^{\frac{1}{n+1}}-1\big)\;\ $ is valid. (I.H.)

So I need to prove that:
$$(n+1)\big(t^{\frac{1}{n+1}}-1\big)\ge (n+2)\big(t^{\frac{1}{n+2}}-1\big)\ $$
But I have not reached anywhere helpful expanding all. Any ideas or different approaches to porve this will be appreciated.
 A: Ok, so continuing with @stity's idea we have two cases:

CASE I: if $\; t\in [0,1)\ \Rightarrow\ t^{\frac{1}{n}}\le t^{\frac{1}{n+1}}\;\ \forall n\in \Bbb N$ 
$$\text{Let}\;\ s=t^{\frac{1}{n(n+1)}}\;\ \forall n\in \Bbb N\; \Rightarrow\ s\le 1$$
Next, let$\ f(x)=(n+1)x^n - nx^{n+1}-1\;\;\ \forall x\in [0,1)\;\ \forall n\in \Bbb N$
$$\Rightarrow\ f'(x)=n(n+1)x^{n-1}-n(n+1)x^n=n(n+1)(x^{n-1}-x^n)\ge 0\;\;\ \forall x\in [0,1)\;\ \forall n\in \Bbb N\\
\text{because}\; x^{n-1}\ge x^{n}\;\ \forall x\in [0,1)\;\ \forall n\in \Bbb N$$
thus $\; f(x)$ is increasing $\;\ \forall x\in [0,1)\; \forall n\in \Bbb N\; $ and, since $s\le 1\; \Rightarrow\ f(s)\le f(1),\;$ i.e.
$$(n+1)t^{1/n+1}-nt^{1/n}-1=(n+1)s^n - ns^{n+1}-1=f(s)\le f(1)=n+1-n-1=0\;\ \forall n\in \Bbb N$$
$$\Leftrightarrow\ (n+1)t^{1/n+1}-nt^{1/n}-1\le0\;\ \forall n\in \Bbb N\\
\Leftrightarrow\ (n+1)t^{1/n+1}-1\le nt^{1/n}\;\ \forall n\in \Bbb N\\
\Leftrightarrow\ (n+1)t^{1/n+1}-1-n\le nt^{1/n}-n\;\ \forall n\in \Bbb N\\
\Leftrightarrow\ (n+1)t^{1/n+1}-(n+1)\le n(t^{1/n}-1)\;\ \forall n\in \Bbb N\\
\Leftrightarrow\ (n+1)(t^{1/n+1}-1)\le n(t^{1/n}-1)\;\ \forall n\in \Bbb N$$
thus $\; \forall t\in [0,1),\;\ n(t^{1/n}-1)\ge (n+1)(t^{1/n+1}-1)\;\ \forall n\in \Bbb N$

CASE II: if $\;t\ge1\ \Rightarrow\ t^{\frac{1}{n+1}}\le t^{\frac{1}{n}}\;\ \forall n\in \Bbb N$, if $t\ge 1$, then
$$\text{Let}\; s=t^{\frac{1}{n(n+1)}}\;\ \forall n\in \Bbb N\; \Rightarrow\ s\ge 1$$
Next, let $\ f(x)=(n+1)x^n - nx^{n+1}-1\;\;\ \forall x\ge 1\;\ \forall n\in \Bbb N$
$$\Rightarrow\ f'(x)=n(n+1)x^{n-1}-n(n+1)x^n=n(n+1)(x^{n-1}-x^n)\le 0\;\ \forall x\ge 1\;\ \forall n\in \Bbb N\\
\text{because}\; x^{n-1}\le x^{n}\;\ \forall x\ge 1\;\ \forall n\in \Bbb N$$
thus, $\; f(x)$ is decreasing $\;\ \forall x\ge 1\;\ \forall n\in \Bbb N$ and, since $s\ge 1\; \Rightarrow\ f(s)\le f(1)$, i.e.
$$(n+1)t^{1/n+1}-nt^{1/n}-1=(n+1)s^n - ns^{n+1}-1=f(s)\le f(1)=n+1-n-1=0\\
\Rightarrow\ (n+1)t^{1/n+1}-1\le nt^{1/n}\;\ \forall n\in \Bbb N\\
\Leftrightarrow\ (n+1)t^{1/n+1}-1-n\le nt^{1/n}-n\;\ \forall n\in \Bbb N\\
\Leftrightarrow\ (n+1)t^{1/n+1}-(n+1)\le nt^{1/n}-n\;\ \forall n\in \Bbb N\\
\Leftrightarrow\ (n+1)(t^{1/n+1}-1)\le n(t^{1/n}-1)\;\ \forall n\in \Bbb N$$
thus $\; \forall t\ge 1,\;\ n(t^{1/n}-1)\ge (n+1)(t^{1/n+1}-1)\;\ \forall n\in \Bbb N$
So, in any case, $\forall t\ge0,\;\; n(t^{1/n}-1)\ge (n+1)(t^{1/n+1}-1)\;\ \forall n\in \Bbb N,\;\ $ i.e. $\big\{n(t^{1/n}-1)\big\}_{n\in \Bbb N}$ is non-increasing
A: It's a very standard result. If $a > 1 > b > 0$ and $r > s > 0$ then it is easy to show that $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s},\,\,\frac{1 - b^{r}}{r} < \frac{1 - b^{s}}{s}\tag{1}$$ The inequality is proved first when $r, s$ are integers, then extended easily to the case when $r, s$ are rationals and finally to the case when $r, s$ are real numbers.
Your question requires only the case when $r, s$ are rational. The problem is solved if we put $a = t$ or $b = t$ and $r = 1/n, s = 1/(n + 1)$. The proof for $(1)$ is easy. Let's start with the case when $r, s$ are positive integers.
Clearly we have $$a^{i} < a^{r}$$ for $i = 0, 1, 2, \dots, r - 1$ and hence on adding these equations we get $$1 + a + a^{2} + \dots + a^{r - 1} < ra^{r}$$ Multiplying the above equation by $(a - 1) > 0$ we get $$a^{r} - 1 < ra^{r + 1} - ra^{r}$$ or $$(r + 1)a^{r} - 1 < ra^{r + 1}$$ Subtracting $r$ from both sides of the above equation we get $$(r + 1)(a^{r} - 1) < r(a^{r + 1} - 1)$$ Diving the above equation by $r(r + 1)$ we get $$\frac{a^{r} - 1}{r} < \frac{a^{r + 1} - 1}{r + 1}\tag{2}$$ Thus the ratio $(a^{r} - 1)/r$ increases as $r$ increases. First part of equation $(1)$ is proved now for the case when $r, s$ are positive integers. The second part related to $b$ can be proved in similar manner starting with $b^{i} > b^{r}$ for $i = 0, 1, 2, \dots, r - 1$.
To establish these inequalities when $r, s$ are positive rationals is easy. Let $r = p/q, s = m/n$ where $m, n, p, q$ are positive integers. Now $r > s$ implies that $pn > qm$. Let $c = a^{1/qn}$ so that $c > 1$. From the inequality $(1)$ established for positive integers $r, s$ it follows that $$\frac{c^{pn} - 1}{pn} > \frac{c^{qm} - 1}{qm}$$ or $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s}$$ so that inequality $(1)$ holds when $r, s$ are rational.
