Prove the inequality $e^x \geq 1+\ln(x+1)$ where $x\geq-1$

Question

How can I prove $e^x\geq1+\ln(x+1)$ for $x>-1$? I was thinking about Bernoulli's inequality or derivatives, but I cannot move forward. Will be very thankful for help.

Answer 1

As it is, the problem is ill-posed since for the value $x=-1$ we can't define $\ln(x+1)$. But if we allow $x\gt-1$ it is valid.

A direct approach:

Let $f(x)=e^x-\ln(x+1)-1$ for $x\in (-1, +\infty]$.

It is easy to see that $f$ is differentiable.

Now we are trying to discern the monotonicity of $f$ by examining $f'$.

We have: $$f'(x)=e^x-\frac1{x+1}$$

Since we can't obtain a direct result we further differentiate $f'$.

$$f''(x)=e^x+\frac1{(x+1)^2}\gt0 \forall x \in (-1, +\infty]$$

So, $f'$ is strictly increasing $\forall x \in (-1, +\infty]$

Thus if $x\in (-1,0]\Rightarrow f'(x)\lt f'(0)=0$.

So $f$ is strictly decreasing for $x\in (-1,0]$.

Similarly we can show that f is strictly increasing for $x\in [0, +\infty]$.

So $\forall x \in (-1,0]\Rightarrow f(x)>f(0)=0$ and $\forall x \in [0,+\infty]\Rightarrow f(x)>f(0)=0$ which provides the result.

Answer 2

$$e^x-x-1$$

has a single minimum at $x=0$, with value $0$. So

$$e^x\ge x+1$$

Then taking the logarithm (a strictly growing function), for $x+1>0$

$$x\ge\log(x+1)$$

and combining,

$$e^x\ge x+1\ge\log(x+1)+1.$$