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Letting $a$ be a strictly positive parameter, one can use calculus to show that $x^a < \frac{1}{1+a-ax}$ for $x \in [0,1)$. Is there a more direct way of showing this? Perhaps some well known inequality that I am not aware off? Thanks for any tips.

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    $\begingroup$ @Asher2211 $0 \le x < 1$ so $0 \le ax < a$ and $a-ax \ge 0$ so $1+ a-ax \ge 1$ and and $0 < \frac 1{1+a-ax} \le 1$. $\endgroup$ – fleablood Jun 18 at 18:47
  • $\begingroup$ @Asher2211: no, because $\frac{1}{1+a-ax} = \frac{1}{1+a(1-x)}$ and $1+a(1-x) > a(1-x) > 0$ given that $a$ is strictly positive and $x <1$. $\endgroup$ – Andres Jun 18 at 18:49
  • $\begingroup$ If $a$ is a natural number then if $x = 1-w$ then $x^a = (1-w)^a = 1 - aw +\text{stuff that oscilates between positive and negative but shrinks in absolute value size so the sum is ultimately positive} > 1+aw = 1 + a(1-x) = 1 +a -ax$. Now just wave your hands about like a banshee and fudge and hem why it can expand to the rational values replacing $x$ and $a=\frac mn$ with $x^{\frac 1n}$ and raising it to the $m$ power and then mutter something about taking limits. $\endgroup$ – fleablood Jun 18 at 18:54
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$$x^a(1+a-ax)=x\cdot x\cdot\dots x(1+a-ax)\leq \left(\dfrac{1+a-ax+ax}{1+a}\right)^{a+1} = 1$$ by simple AM-GM, if $a$ was a positive integer. But then again, you can have weighted AM-GM for real numbers so you can modify this to work for positive real $a.$ But then again, weighted AM-GM, or in general the power-mean equality requires some calculus in its proofs so I am not sure if you can make things any simpler.

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