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The following post is related to If $x \in [0,1]$ then how do you show the inequality $r > s > 0$ implies $x^r < x^s$? and can be thought of as a generalization to the questions posted previously. The backround for these problems comes from the need to rule out cases of fractional powers when the inequality below is reversed in (particular $x^s > y^r$). The question can be formulated as follows

If $ 0< x < y$ and $0 < s < r$ then when is the following inequality true $x^s < y^r$?

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2 Answers 2

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If $y>x>1$ and $r>s>0$ then $x^r<y^s$ holds good because $y^s>y^r$ and $y^r>x^r$ therefore $y^s>x^r$

If $1>y>x>0$ and $r>s>0$ then $x^r<y^s$ doesn't good because $0.25<0.5$ and $0.5<2$ but $0.25^{0.5}\not< 0.5^{2}$

If $y>1$, $1>x>0$ and $r>s>0$ then $x^r<y^s$ holds good because $x^r<1$ $\forall r\in \mathbb{R^+}$ and $x\in (0,1)$ , $y^r>1$ $\forall s \in \mathbb{R^+} $ and $y>1$

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@user7980: Is this what you are looking for???? –  Ramana Venkata Oct 4 '11 at 9:16

No! Consider the case when $x$ and $y$ are below $1$ and $r$ is very large.

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