Q.Prove Bernoulli's inequality: $(1+x)^n \ge 1 + nx$ for every natural number $n$, where $x > -1$.

[!TLDR] Multiply the inductive hypothesis by $(1+x)>0$ and drop the non-negative term $kx^2$. [!ANSWER] Proved: $(1+x)^ …

Let $P(n)$: $(1+x)^n\ge1+nx$, for fixed $x>-1$. Base case: $n=1$ gives $(1+x)^1=1+x=1+1\cdot x$, so equality holds and $P(1)$ is true. Inductive step: assume $P(k)$: $(1+x)^k\ge1+kx$ for some $k\ge1$. Since $x>-1$, we have $1+x>0$, so multiplying both sides of the inductive hypothesis by $(1+x)$ preserves the inequality direction: $(1+x)^{k+1}=(1+x)^k(1+x)\ge(1+kx)(1+x)=1+x+kx+kx^2=1+(k+1)x+kx^2$. Since $k\ge1$ and $x^2\ge0$, we have $kx^2\ge0$, so $ …

Inequality-by-induction: multiply through by the positive quantity $(1+x)$, then discar …

Multiplying an inequality by $(1+x)$ without first checking $1+x>0$ -- this is exactly why the hypothesis $x>-1$ is required; multipl …