e^x ≥ 1 + x

The easiest way to see this is to notice that \(e^x\) is convex (second derivative \(=e^x > 0\)). So any tangent line will lie below the curve. The tangent at \(x=0\) is \(1+x\).

copied from this stack overflow answer

Start with definition of the exponential function: \[ e^x = \lim_{n\rightarrow \infty} \left(1 + \frac{x}{n}\right)^n \] We'll show that the sequence \(e_n(x) = \left(1+\frac{x}{n}\right)^n\) is monotonically increasing for \(x>-1\). Let's look at this ratio:

\[\begin{align} \frac{e_{n+1}(x)}{e_{n}(x)} &= \frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^n}\\ &= \left(1 + \frac{-x}{(x+n)(n+1)} \right)^{n+1} \left(1 + \frac{x}{n}\right)\\ &\geq \left(1 + \frac{-x}{(x+n)} \right)\left( 1 + \frac{x}{n} \right)\\ &= 1 \end{align}\]

To get from (1) to (2) is arithmetic. To get from (2) to (3) is Bernoulli's inequality which says that \((1+y)^r \geq 1 + ry\) for any integer \(r\geq 0\) and any real number \(y \geq -1\). We know that these conditions hold because \(n \geq 0\) and \(x \geq -1\), and \(-n \leq x + 1 \Leftrightarrow \frac{-x}{(x+n)(n+1)} \geq -1\).

Then, we've shown that \(e^x\) is the upper bound of a montonically increasing sequence. So it is \(\geq\) than each item in that sequence. In particular, we have \[ e^x \geq 1 + x \] for all \(x > -1\). For all \(x \leq -1\), note that we also have \(e^x \geq 1 + x\), since \(e^x\) is always positive. So we have \(e^x \geq 1+x\) for all \(x\).

• \(x \geq \ln(1+x)\) – from taking \(\ln\) of both sides and using the fact that \(\ln\) is monotonically increasing.
• \(u - 1 \geq \ln(u)\) – from \(u = 1 + x\)
  • Aside: when can we use change of variables? See this stack overflow question.
  • Basically, we can replace every \(x\) with an \(f(x)\) provided that the range of \(f(x)\) is in the domain of the original equation
    • Because the original equation is telling us some fact about \(x\). And we can replace that \(x\) with any old complicated procedure, as long as we do it in every place that \(x\) is mentioned
• \(-u \geq \ln(1-u)\) from \(u = -x\)