$\lvert \exp(x+h) - \exp(x) \rvert \le \lvert h \rvert \exp( \lvert h \rvert)\exp(x)$

Question

Prove that

$$\lvert \exp(x+h) - \exp(x)\rvert \le \lvert h \rvert \exp( \lvert h \rvert)\exp(x)$$

$\forall x, h \in \mathbb{R}$.

First, I thought about using induction here, but wasn't sure if this would work, since $\exp(x+h)$ and $\exp( \lvert h \rvert)$ might have a different sum control variable. Unfortunately I don't see, where I could intelligently estimate and probably use the triangle inequality to form the expression to a correct proof. Maybe someone can help me with that. Thanks in advance !

Answer 1

For $h\ge 0$, the claim amounts to $$ \exp(x+h)-\exp(x)\le h\exp(h)\exp(x),$$ which, after dividing by $\exp(h)\exp(x)=\exp(x+h)$, is equivalent to $$1-\exp(-h)\le h, $$ or $$ 1-h\le \exp(-h),$$ an instance of the well-known inequality $$\tag1 \exp(t)\ge 1+t.$$

For $h=-k<0$, the claim amounts to $$\exp(x)-\exp(x-k)\le k\exp(k)\exp(x),$$ which, after dividing by $\exp(x)$, is equivalent to $$ 1-\exp(-k)\le k\exp(k)$$ Now note that by $(1)$ $$ k\exp(k)\ge k(1+k)>k\ge1-\exp(-k).$$

Answer 2

$$\exp(x+h)-\exp x= \exp x(\exp h-1)$$

But $$|e^h-1| = \left||\int_0^h e^t dt\right|\le \left|e^h\int_0^hdt\right| = |he^h|$$