Alternative proofs $\sqrt[\alpha]{2} < \frac{\alpha}{\alpha-1}$ I am looking for smart proofs of the inequality 
$$\sqrt[\alpha]{2} < \frac{\alpha}{\alpha-1} \qquad \text{for} \ \alpha> 1$$
this inequality comes from the comparison of the median and the mean of a $\mathrm{Pareto(\alpha)}$ distribution.
Here is the proof I found but I would like a list of different ways to approach the problem.
The inequality rewrites as 
$$ 2 < \left( 1+\frac{1}{\alpha-1} \right)^\alpha $$
the RHS expands (Newton generalized) as 
$$ 1 + \alpha \frac{1}{\alpha- 1} + \frac{\alpha (\alpha -1)}{2}\left (\frac{1}{\alpha- 1} \right)^2 + \frac{\alpha (\alpha -1)(\alpha-2)}{3!}\left (\frac{1}{\alpha- 1} \right)^3 + \dots$$
and thanks to Leibnitz criterion (alternating signs might begin at the fourth term) this is greater than the fist two terms which are greater than 2.
What I am not satisfied with is that it looks quite a long proof (actually, despite is trivial, the case $\alpha \in (1,2)$ should be treated separately) and indeed the inequality can be stronger substituting 2 with $e$.
What I would like is a (let's say) creative proof rather than simply checking that the limit at $+\infty$ is $e$ and showing that the function is decreasing for $\alpha>1$.
 A: $f(x)=2^x$ is a convex function on $[0,1]$, hence its graph lies above any tangent and below the secant line through $(0,1)$ and $(1,2)$, namely $y=x+1$. It follows that
$$ \forall \alpha>1,\qquad 2^{1/\alpha} < 1+\frac{1}{\alpha} $$
that is stronger than the given inequality, since 
$$ \frac{\alpha}{\alpha-1} = \frac{1}{1-\frac{1}{\alpha}} = 1+\frac{1}{\alpha}+\frac{1}{\alpha^2}+\frac{1}{\alpha^3}+\ldots $$
A: Using Bernoulli inequality and $\dfrac 1\alpha<\dfrac 1{\alpha-1}$ we get
$$\sqrt[\alpha]{2} = (1+1)^{\frac 1\alpha} \le 1 + \frac 1 \alpha < 1 + \frac{1}{\alpha-1} = \frac{\alpha}{\alpha-1}.$$
A: Using $\log (1+x) \le x$ one gets
$$
 \log \frac{\alpha}{\alpha -1} = - \log \left( 1 - \frac 1 \alpha \right) 
 \ge  \frac 1 \alpha
$$
and therefore 
$$
2 ^{\frac 1 \alpha} < e^{\frac 1 \alpha} \le \frac{\alpha}{\alpha -1}
$$
A: For $f(x)=\ln x$ in $[\alpha-1,\alpha]$ is continuous and diferentiable  ($\alpha>1$), with Mean Value theorem in this interval we have $\alpha-1<\zeta<\alpha$ such that
$$\ln(\alpha)-\ln(\alpha-1)=\frac{1}{\zeta}>\frac{1}{\alpha}$$
then
$$\left(\frac{\alpha}{\alpha-1}\right)^\alpha>e>2$$
this concludes what we want.
