# The inequality $(ae^{bx} - be^{ax} - (a -b)) > 0$ [closed]

I'm struggling to see why for $$0 it's true that for every $$x\in \mathbb{R}, x>0$$ : $$(ae^{bx} - be^{ax} - (a -b)) > 0$$

We can use Jensen's Inequality:

Let $$\displaystyle f(z)= e^{zx}$$ and as $$f(z)$$ is convex, implies: $$\ \displaystyle ae^{bx} - be^{ax} \geq (a-b) f \left ( \frac{ab-ba}{a-b}\right ) = (a-b) \$$

Therefore,

$$\displaystyle ae^{bx} - be^{ax} - (a-b) \geq 0 .$$

Here, there is absolutely no restriction on $$a,b$$. And the equality holds when the condtion $$a=b$$ is satisfied.

• Nice answer. A minor point is showing the case(s) for equality based on the question as posed by OP. Mar 13, 2022 at 16:03
• @JacobA oh ok, Will Edit it. Thank you for the suggestion. Mar 13, 2022 at 16:12
• Just wondering: $\frac{a}{a-b}$ and $-\frac{b}{a-b}$ play the role of the weights here as far as I understand, but the weights should be non-negative, or? Mar 13, 2022 at 19:19
• @Diger By the way, equivalently, you may use $\frac{b-a}{b}f(0) + \frac{a}{b}f(b) \ge f(a)$ which is $\frac{b-a}{b} + \frac{a}{b}\mathrm{e}^{bx} \ge \mathrm{e}^{ax}$. Mar 14, 2022 at 1:08

Consider the function $$f(x)=ae^{bx}-be^{ax}$$ and note that $$f(0)=a-b$$. Next note that $$f'(x)=abe^{bx}-abe^{ax}=abe^{ax}(e^{(b-a)x}-1)$$ which vanishes only for $$x=0$$. Can you finish?

The inequality holds for $$x\ne0$$, not only for $$x>0$$.

• you're right, I wrote the question wrong, I corrected it Mar 13, 2022 at 16:11