Bernoulli's inequality, \((1+x)^n \ge 1+nx\), is one of the workhorses of elementary analysis, but it is classically stated only for a base of the form \(1+x\) and only for \(x \ge -1\). This page explores a generalization that moves two things at once: it puts a translation parameter \(y \ge 1\) into the base and lets the exponent \(z \ge 1\) be any real number, which widens the domain of \(x\) from \([-1,\infty)\) to \([-y,\infty)\). Move the parameters below and watch the curve, its linear lower bound, and the gap between them.
- Curve \((y+x)^z\)
- Bound \(y+xz\)
- Gap
- Where it fails
- Auxiliary \(f(t)=t^z-tz\)
- Minimum \(1-z\)
- Lower bound \(y(1-z)\)
- Margin above the minimum
The result
Theorem (Bernoulli with translation and real exponent)
For all \(y, x, z \in \mathbb{R}\) with \(z \ge 1\), \(y \ge 1\) and \(x \ge -y\):
\[(y+x)^z \ \ge\ y + xz\]
Equality holds if and only if \(x = 0\) or \(z = 1\).
Proof
Put \(t = x + y\). The hypothesis \(x \ge -y\) gives \(t \ge 0\), and the statement becomes \(t^z \ge tz + y(1-z)\), so it suffices to prove that \(f(t) := t^z - tz \ge y(1-z)\).
Step 1: the global minimum of \(f\)
Differentiating, \(f'(t) = z(t^{z-1}-1)\) and \(f''(t) = z(z-1)t^{z-2}\). Since \(z \ge 1\) and \(t > 0\), we get \(f'' \ge 0\), so \(f\) is convex on \((0,\infty)\). Its only critical point there is \(t = 1\), which convexity turns into a global minimum: \(f(t) \ge f(1) = 1-z\). The bound extends to \(t = 0\) by continuity, since \(f(0) = 0 \ge 1-z\).
Step 2: comparison with \(y(1-z)\)
Because \(z \ge 1\) we have \(1-z \le 0\), and because \(y \ge 1\), multiplying by \(y\) only makes it smaller: \(y(1-z) \le 1-z\). Chaining both steps gives \(y(1-z) \le 1-z \le f(t)\), which is exactly what was needed.
The equality case
Equality requires both steps to be tight at once. Step 1 is tight only at \(t = 1\), that is \(x = 1-y\); step 2 only when \(y = 1\) or \(z = 1\). If \(z = 1\) the statement collapses to the identity \(y+x \ge y+x\). If \(z > 1\), the two conditions together force \(y = 1\) and hence \(x = 0\). So equality holds exactly when \(x = 0\) or \(z = 1\).
Corollary
Setting \(y = 1\) recovers the standard real-exponent form \((1+x)^r \ge 1+rx\) for \(r \ge 1\) and \(x \ge -1\); restricting further to a natural exponent gives the classical inequality. Press Classic above to land on that case.
Why \(y \ge 1\) cannot be dropped
If \(0 < y < 1\) and \(z > 1\), the comparison in step 2 reverses and the conclusion can fail. Take \(y = \tfrac{1}{2}\), \(z = 2\), \(x = 0\): the left side is \(\tfrac{1}{4}\) while the right side is \(\tfrac{1}{2}\). Press Counterexample above and the failing region shows up in red.
Against earlier generalizations
Classical extensions to a real exponent keep the restriction \(x \ge -1\). Tying the lower end of the domain to the translation parameter is what widens it to \(x \ge -y\).
| Result | Hypotheses | Inequality |
|---|---|---|
| classic | \(n \in \mathbb{N},\ x \ge -1\) | \((1+x)^n \ge 1+nx\) |
| realExp | \(r \ge 1,\ x \ge -1\) | \((1+x)^r \ge 1+rx\) |
| weierstrass | \(x_i > -1\) | \(\prod (1+x_i) \ge 1 + \sum x_i\) |
| refined | \(r \ge 1,\ x \ge -1\) | \((1+x)^r \ge 1+rx+\tfrac{r(r-1)}{2}x^2\) |
| thisWork | \(z,\ y \ge 1,\ x \ge -y\) | \((y+x)^z \ge y+xz\) |
Where it is useful
analysis
analysis
probability
probability
convex
convex
finance
finance
Full write-up
The complete article, with the formal proof, the comparison against the literature and every figure, is available as a PDF.