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


2 Answers 2


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) \ $


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

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

  • $\begingroup$ Nice answer. A minor point is showing the case(s) for equality based on the question as posed by OP. $\endgroup$
    – Jacob A
    Mar 13, 2022 at 16:03
  • $\begingroup$ @JacobA oh ok, Will Edit it. Thank you for the suggestion. $\endgroup$ Mar 13, 2022 at 16:12
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    $\begingroup$ 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? $\endgroup$
    – Diger
    Mar 13, 2022 at 19:19
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    $\begingroup$ @Diger By the way, equivalently, you may use $\frac{b-a}{b}f(0) + \frac{a}{b}f(b) \ge f(a)$ which is $\frac{b-a}{b} + \frac{a}{b}\mathrm{e}^{bx} \ge \mathrm{e}^{ax}$. $\endgroup$
    – River Li
    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 $x>0$.

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

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