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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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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0I can see a relation with "number theory"... anyone know a simple approach that don't require derivative? – 2012-04-23