Proof e^{a v} (a (v-w)-1)+e^{a w} is positive if v > w and a > 0

[BJU]

Proof [tex]e^{a v} (a (v-w)-1)+e^{a w}[/tex] is positive if v > w and a > 0

This is obviously related to my other post, but seems to be a separate problem.

I have the term

\(\displaystyle e^{a v} (a (v-w)-1)+e^{a w}\)

and (in a derivative) a similar term

\(\displaystyle e^{a v} (a (v-w)-2)+e^{a w} (a (v-w)+2)\)

with v > w and a > 0.

I have the strong intuition that both terms are positive, but I have no clue how to prove this. \(\displaystyle - e^{a v} +e^{a w}\) is negative, so why is \(\displaystyle e^{a v} (a (v-w))\) always making up for it? I just don't see it.

I thought I could somehow use that the terms are 0 if a=0 and then use the derivative, but the derivative contains similar expressions that I would have to prove again. It never ends.

Does anyone have any hints?

Edit: I guess this is an Algebra question?

 

[BJU]

Ok, the fact that I continue to solve my own problems probably does reflect badly on me, but somehow posting the problems helps me think differently about them.

Anyway, trying to solve this by looking at a may not be the way to go.

But if we start at v=w, the \(\displaystyle e^{a v} (a (v-w)-1)+e^{a w}=0\).

The partial derivative of the term with respect to v is

\(\displaystyle a^2 e^{a v} (v-w) > 0\) for any a>0.

Hence, the original expression is positive for all a>0, v>w.

QED?

 

[Deleted member 4993]

Ok, the fact that I continue to solve my own problems probably does reflect badly on me, but somehow posting the problems helps me think differently about them.

No... it shows that you are serious about mathematics and you are willing to share your knowledge/experience.

A very commendable trait...

 

[Ishuda]

This is obviously related to my other post, but seems to be a separate problem.

I have the term

\(\displaystyle e^{a v} (a (v-w)-1)+e^{a w}\)

and (in a derivative) a similar term

\(\displaystyle e^{a v} (a (v-w)-2)+e^{a w} (a (v-w)+2)\)

with v > w and a > 0.

I have the strong intuition that both terms are positive, but I have no clue how to prove this. \(\displaystyle - e^{a v} +e^{a w}\) is negative, so why is \(\displaystyle e^{a v} (a (v-w))\) always making up for it? I just don't see it.

I thought I could somehow use that the terms are 0 if a=0 and then use the derivative, but the derivative contains similar expressions that I would have to prove again. It never ends.

Does anyone have any hints?

Edit: I guess this is an Algebra question?

If we keep it at the calculus level, we have
\(\displaystyle e^{a v} (a (v-w)-1)+e^{a w}\, =\, e^{a w}\, f(t)\)
where
\(\displaystyle t\, =\, a\, (v-w)\)
and
\(\displaystyle f(t) = [e^t\, (t-1)+1]\)

Now, obviously, the sign of the original expression with its conditions depends on f(t) for t>0.

 

[BJU]

If we keep it at the calculus level, we have
\(\displaystyle e^{a v} (a (v-w)-1)+e^{a w}\, =\, e^{a w}\, f(t)\)
where
\(\displaystyle t\, =\, a\, (v-w)\)
and
\(\displaystyle f(t) = [e^t\, (t-1)+1]\)

Now, obviously, the sign of the original expression with its conditions depends on f(t) for t>0.

I needed a bit to wrap my head around this, but this is ingenious. This helps me with some of my other stuff, too.

In hindsight, I feel like I should have been able to come up with this myself, but my last math lecture was more than a decade ago (boy, am I old) and some things are a bit slow to come back.

Thanks very much for the support!

 

[BJU]

\(\displaystyle f(t) = [e^t\, (t-1)+1]\)

Now, obviously, the sign of the original expression with its conditions depends on f(t) for t>0.

So since \(\displaystyle f(0)=0\) and \(\displaystyle f'(t)=t e^t\), the expression is always positive for t>0.

Or is there an even simpler way to prove this?