Proving $x^\alpha-\alpha x \le 1- \alpha $ 
Prove: $$x^\alpha-\alpha x \le 1- \alpha  \\
\forall x\ge 0 \ ,  \ 0<\alpha <1$$

It does resamble Lagrange's MVT, in order to get to the RHS, say $\alpha\in[0,1]\Rightarrow \frac{1-\alpha-0^\alpha-\alpha 0}{1-0}=1-\alpha$.
But I'm not sure what to do about the LHS. I always get a negative value in the LHS and we know that the RHS will always be postive but that doesn't really help with proving it...
Edit: maybe define a function: $g(x) = x^\alpha-\alpha x - 1+ \alpha$ Derive it: $g'(x) = \alpha x^{\alpha-1}-\alpha $ but then I get to $x^{\alpha-1}=0$

Also, I had two questions when trying to figure this out:


*

*Can we use deriviation of both sides of an inequality to prove it ?

*Can we use limit of both sides of an inequaltiy to prove it ?
Thanks.
 A: Let $f(x)=x^\alpha-\alpha x-\alpha+1$ for $x\geq 0$, where $\alpha<1$, so $f'(x)=\alpha x^{\alpha-1}-\alpha$ and $f''(x)=\alpha(\alpha-1)x^{\alpha-2}$. Then $f'(x)=0\Rightarrow x=1$ with $f''(1)=\alpha(\alpha-1)\leq 0$. This implies that $f$ has a local maximum at $x=1$ and so $f(1)=2(1-\alpha)\geq1-\alpha=f(0)$ . Also we have $\displaystyle \lim_{x\to\infty}f(x)=-\infty$. Since $f$ is continuous for $x\geq 0$, considering above values we see that $f$ has a maximum at $x=1$ and equals $f(1)=2(1-\alpha)$. Therefore we have $f(x)\leq f(1)$ for all $x\geq 0$ and this leads to the desired inequality.
A: The problem is too easy once one rewrites it as $x^{\alpha} - 1 \leq \alpha(x - 1)$. If $f(x) = x^{\alpha}$, then $f'(x) = \alpha x^{\alpha - 1}$ and then by Mean Value Theorem we have $$x^{\alpha} - 1 = f(x) - f(1) = (x - 1)f'(c) = \alpha(x - 1)c^{\alpha - 1}$$ where $c$ is between $1$ and $x$. Clearly we need to consider cases when $x < 1$ and $x > 1$. For $x = 1$ the result is trivial. Supposing that $x > 1$ we need to ensure that $c^{\alpha - 1} < 1$ which is practically obvious as $1 < c < x$ and $\alpha - 1 < 0$. Same way if $x < 1$ then we need to establish that $c^{\alpha - 1} > 1$ and again this is obvious because then $x < c < 1$.
