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I'm trying to prove (maybe using Lagrange theorem, but with no real success) that when $x \ge 0$ and $0<a<1$ that $$x^a -ax \le 1-a$$ I defined a function to be: $$f(x)=x^a-ax+a-1$$ and therefore: $$f(1)=0 , f(0)=a-1$$ Was also thinking of using the fact that $x^a$ is strictly ascending (but don't really know how)

Thanks for help.

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Your inequality is of the form $g(x) < g(1)$ where $g(x)=x^a-ax$. It does not hold for $x=1$. –  Siméon Jan 15 '13 at 9:00
    
thanx, typo in copying the question fixed. –  user1685224 Jan 15 '13 at 9:06
    
My previous comment contains a hint: show that $g(x)$ attains its maximum at $x=1$. –  Siméon Jan 15 '13 at 9:09

2 Answers 2

$f'(x)=ax^{(a-1)}-a=a(x^{(a-1)}-1)$

$a-1<0$ and $x>0$ and $x<1$ so $x^{(a-1)}>1$

so $f' >0$ and $f(1)$ is the maximum, answering the question.

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Note that $$f'(x)=ax^{a-1}-a=a\left(x^{a-1}-1 \right) \Rightarrow f'(1)=0$$
If $0 < a < 1,$ then for $x>1$ $$x^{a-1}-1<0 \Rightarrow f'(x)<0, $$ for $0 < x < 1$ $$f'(x)>0 .$$ This means that $f(x)$ has maximum at $x=1,$ therefore, $f(x)\leqslant{0} $ for all $x>0$

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