Finding values of b

**mybrohshi5** · November 29th 2010, 07:10 AM

Not sure if this post should go here so if it would be better else where would a Mod please move this

Thanks!

Here is the question i am working on and it seems simple enough but i just wanted a second opinion

For what value(s) of b is $b^a$ > ab for all real numbers a?

if b is > 2 then $b^a$ will be > ab

but

I also notice that if a = 1 then b can also = 1 so i am not sure what to do with that.

Any help on how to prove this problem would be great

Thank you

**tonio** · November 29th 2010, 08:34 AM

Originally Posted by mybrohshi5

Not sure if this post should go here so if it would be better else where would a Mod please move this

Thanks!

Here is the question i am working on and it seems simple enough but i just wanted a second opinion

For what value(s) of b is $b^a$ > ab for all real numbers a?

if b is > 2 then $b^a$ will be > ab

but

I also notice that if a = 1 then b can also = 1 so i am not sure what to do with that.

Any help on how to prove this problem would be great

Thank you

It must be $b>0$ since $b<0\Longrightarrow b^a$ is not defined for many real values of $a$ . If $b=0$ then the

inequality is always true insofar we define $0^0=1$ .

If $a\leq 0$ the inequality is trivially true, as the LHS is positive and the RHS is non positive, so we can assume $a>0$ , then:

$b^a\geq ab\Longleftrightarrow a\ln b\geq \ln a+\ln b\Longrightarrow (a-1)\ln b\geq \ln a$ , and since the ineq. is true for

$a=1$ we can assume $0<a\neq 1$ , so it must be that

1) $a>1\Longrightarrow \displaystyle{\ln b\geq \frac{\ln a}{a-1}\Longrightarrow b\geq a^\frac{1}{a-1}}$ .

But $\displaystyle{\lim\limits_{a\to 1^+}a^\frac{1}{a-1}=\lim\limits_{a\to 1^+}e^\frac{\ln a}{a-1}=e}$ , so it must be $b\geq e$ , and this value already covers the whole

case $a>1$ , as $\displaystyle{a^\frac{1}{a-1}}$ has a maximum at $a=2$

2) $0<a<1\Longrightarrow$ we get that $\displaystyle{a^\frac{1}{a-1}\xrightarrow [a\to 0]{}\infty}$ , so in this case $b$ cannot be bounded by some value.

Tonio

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