trouble with inequality

Inquiry: I intend to address $0<1−an/(mb^2)e^{−r(T−t)}<1$, where $r, a, b, T, t>0$.

The remedy is that either $$an\leq mb^2$$ or $$mb^2\leq an\leq mb^2e^{rT}$$ and also $$t< T − (\ln(an) − \ln(mb^2 ))/r$$.

My Attempt: My ideas are that the first component $0<1−\frac{an}{mb^2}e^{−r(T−t)}$ offers me $an \leq mb^2$ due to the fact that $e^{−r(T−t)}>0$, so I have the first component. The 2nd component $1−\frac{an}{mb^2}e^{−r(T−t)}<1$ does not offer me valuable details given that $\frac{an}{mb^2}e^{−r(T−t)}>0$ constantly.

Just how do I get the various other fifty percent of the remedy ( $mb^2\leq an\leq mb^2e^{rT}$ and also $t< T − (\ln(an) − \ln(mb^2 ))/r$)?

I additionally become aware that the trouble I need to address lowers to addressing $xy<1$ where both $x,y>0$.

Combined from: tricky inequality

How do I deal with addressing $0<1−\frac{an}{mb^2}e^{−r(T−t)}<1$, where a, b, T > t > 0? I have actually been stuck below for time currently.

2019-05-05 06:13:10
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Answers: 1

There are several unneeded variables. Allow

$$\alpha = \frac{an}{mb^2}.\qquad(1)$$

Then the inequality comes to be

$$ 0 < 1 - \alpha e^{-r (T - t)} < 1 $$

The first noticeable action is execute "1 −" on every components,

$$ 1 > \alpha e^{-r (T-t)} > 0 $$

Since the rapid function's array declares and also < 1 (given that r > 0 and also T > t > 0), we can make certain α declares.

If 0 < α ≤ 1, after that every t will certainly please the inequality (the first remedy).

So think α > 1. Currently it's rather noticeable on just how to address t in regards to r , T and also α . Replacement (1) once more to come back a , n , m and also b .

Points you might take into consideration :

  • e x is purely raising.
2019-05-08 20:50:34