How to prove $x^{y-1}\geq xy$ with $x,y\in \mathbb{R}$ with $x,y\geq 3$ . Do I need induction? Or is there an elegant way?
inequality proof of $x^{y-1} \ge xy$
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number-theory
inequality
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1Number theory? ${}{}$ – 2012-04-23
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0I get this problem in an number theory course – 2012-04-23
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4Induction works for natural numbers, which in this particular inequality is not the case. – 2012-04-23
1 Answers
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$$x^{y-2} \geq 3^{y-2} \geq y,\ \forall x,y \geq 3$$ where the last inequality can be proven by induction in a very simple way for $y \geq 3$ integer.
For $y \geq 3$ you can use the function $f(y)=3^{y-2}-y$ and show that is increasing on $[3,\infty)$.
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2I can't see a "very simple way" for doing induction over $\mathbb{R}$. Maybe the question was over $\mathbb{N}$, or you need to show that $f(y)=3^{y-2}-y$ is increasing for $y\geq3$ (that is true, and can be done with derivative). – 2012-04-23
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0Sorry. I forgot that $y \geq 3$ was real. – 2012-04-23
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0I have to proof: $x^{y-1}\geq xy$ for all $x,y\in\mathbb{R}$ and $x,y\ge 3$ – 2012-04-23
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0@evenpoly: Ok, but can't you see the connection? – 2012-04-23
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0Ok I see it. The derivative of ${3}^{y-2}-y$ is $\frac{d}{dy}3^{y-1}-y={3}^{y-2}\ln \left( 3 \right) -1$. And this is greater than 0 for y>2. So we have $3^{y-1}-y > 0$ for $y>3$ and the function $3^{y-1}-y$ is monotonically nondecreasing for y>3. So the inequality holds. Is that ok? – 2012-04-23
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0@evenpoly: Yes, that's ok. – 2012-04-23
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0Great. Thank you all for your help. My idea of induction was really worse since the situations takes place in $\mathbb{R}$. – 2012-04-23
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0I can see a relation with "number theory"... anyone know a simple approach that don't require derivative? – 2012-04-23