Proof for $f(x)=e^x+x$ strictly increasing

Question

I am asked:

Let $f:\mathbb{R} \to \mathbb{R}$ be given by $f(x)=e^x+x$.

Prove that $f$ is strictly increasing. That is to say, prove that for any $a,b\in \mathbb{R} $ $$(a<b)\implies (e^{a}+a<e^{b}+b)$$

Here's my attempt:

For any $a,b\in \mathbb{R}$, assume that $a<b$, i.e. $b=a+c$ for some $c\in \mathbb{R}^+$.

Then, $e^a+a<e^b+b\implies e^a+a<e^{a+c}+(a+c)\implies e^a+a<e^a*e^c+a+c\implies e^a<e^a*e^c+c\implies 1<e^c+\frac{c}{e^a}\implies 0<e^c+\frac{c}{e^a}-1$.

Since $c\in \mathbb{R}^+,e^c>1$. Therefore, $0<e^c+c$ is true for $c\in \mathbb{R}^+$. Thus, the original implication is true, so the function $f$ is strictly increasing. $\blacksquare$

Did I miss anything? Thanks!

Answer 1

Your reasoning is right besides one crucial detail.

You say

Therefore, $0<e^c+c$ is true for $c\in \mathbb{R}^+$. Thus, the original implication is true, so the function $f$ is strictly increasing.

But when you show $A \Longrightarrow B$ and show that $B$ holds, you don't know anything about $A$. To show that $A$ holds, you have to show $B\Longrightarrow A$, so in your reasoning (but not your initial statement you want to prove) all "$\Longrightarrow$" must be "$\Longleftarrow$" instead. Luckily, as you can easily verify, it is even "$\Longleftrightarrow$". For example

$$1<e^c+\frac{c}{e^a}\iff 0<e^c+\frac{c}{e^a}-1$$

because, as you explained, $c\in\mathbb{R}^+$

Answer 2

You can prove directly (without using $c$): $$a<b \Rightarrow \begin{cases}e^a<e^b \\ a<b\end{cases} \stackrel{+}{\Rightarrow} e^a+a<e^b+b.$$ In general, the sum of two increasing function is also an increasing function.