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suppose we are given real numbers $x,y$ and positive integers $m,n$. is it true that

$$ mx + ny \leq x^m + y^n \leq (xy)^{m+n} $$

???

i would like someone to tell me what this inequality means geometrically, and how can we solve it algebraically? thanks.

however it seems it does not work if $x,y$ are less than 1. so let us put condition that both should be greater than 1

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    $\begingroup$ If $x = y = 1$, the far right hand side equals $1$ and the middle equals $2$ no matter what $m$ and $n$ are. Also no matter what $m$ and $n$ are, by continuity, the inequality is still the wrong way for $x = 1 + \epsilon$ and $y = 1 + \delta$ for some positive $\epsilon, \delta$. I think we need to strengthen the assumptions to $x, y \geq 2$ $\endgroup$ – Arthur Dec 5 '13 at 11:57
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Look at the parts individually.

First, show that $$ mu \leq u^m $$ for $u \geq 2$ (you need to require that for your theorem to hold, as the comments point out), and arbitrary $m \in \mathbb{N}$. To show that, observe that it holds for $m=0$ and $m=1$, and then use induction on $m$. From that, the first inequality follows, because if $a' \leq a$ and $b' \leq b$ then $a' + b' \leq a + b$.

For the second part, distinguish between the cases $x^m \leq y^n$ and $x_m > y^n$. In the first case, you have $x^m < y^n \leq 2y^n$. Now show that $2y^n \leq (xy)^{m+n}$ (use again that $x,y \geq 2$, and induction on $m$). Together you get $$ x^m < y^n \leq 2y^n \leq (xy)^{m+n} $$ which proves the second inequality. The second case works the same, except that you start out with $x^m < y^n \leq 2x^n$.

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