# inequality proof of $x^{y-1} \ge xy$

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?

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Number theory? ${}{}$ –  user21436 Apr 23 '12 at 15:20
I get this problem in an number theory course –  wieschoo Apr 23 '12 at 15:24
Induction works for natural numbers, which in this particular inequality is not the case. –  Thomas E. Apr 23 '12 at 15:25

$$x^{y-2} \geq 3^{y-2} \geq y,\ \forall x,y \geq 3$$ 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 $[3,\infty)$.
I 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). –  carlop Apr 23 '12 at 15:29
Sorry. I forgot that $y \geq 3$ was real. –  Beni Bogosel Apr 23 '12 at 15:31
I have to proof: $x^{y-1}\geq xy$ for all $x,y\in\mathbb{R}$ and $x,y\ge 3$ –  wieschoo Apr 23 '12 at 15:35
Ok 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? –  wieschoo Apr 23 '12 at 16:09