# $r^a \leq r^b+r^c$ given $a \leq b+c$?

Apologies for the (maybe obvious) question. It's come up as part of a proof of the triangle inequality for a metric function I'd like to work with.

For real number $r \geq 1$ and positive integers a, b, c:

Is it generally true that $r^a \leq r^b+r^c$? Given that $a \leq b + c$.

Happy to accept a pointer in the right direction.

Thanks

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3
have a look at $r=3$, $a=2$, $b=c=1$. – Alexander Thumm Mar 29 '11 at 12:01
well I feel somewhat stupid! Much obliged. – Tom Seaton Mar 29 '11 at 12:08

## 1 Answer

Since $x\mapsto r^x$ is nondecreasing when $r\geq 1$, you know that $r^a\leq r^{b+c}$, and nothing sharper than this can hold in general (because $x\mapsto r^x$ is strictly increasing when $r>1$, and $b+c$ might in fact equal $a$). Since $r^{b+c}=r^br^c$, the question amounts to whether $r^br^c\leq r^b+r^c$. Although you're restricting $b$ and $c$ to positive integers, you still should expect this to be false, because a product of positive numbers is not generally smaller than their sum. And in fact, it isn't hard to find counterexamples, one of which was given in a comment by Alexander Thumm.

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