Q.Prove that $(1 + x)^n \ge (1 + nx)$, for all natural number $n$, where $x > -1$.

Let $P(n)$ be the statement: $(1+x)^n\ge1+nx$, for a fixed real number $x>-1$.

Base case: For $n=1$,

so $(1+x)^1\ge1+1\cdot x$ holds with equality. $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$:

We must show $(1+x)^{k+1}\ge1+(k+1)x$.

Since $x>-1$, we have $1+x>0$. Multiplying both sides of the induction hypothesis by the positive number $(1+x)$ preserves the inequality:

Since $k\ge1$ and $x^2\ge0$, we have $kx^2\ge0$, so

$1+(k+1)x+kx^2\ge1+(k+1)x.$ …

Inequality induction where the condition $x>-1$ is essential: it guarantees $1+x>0$, so multiplying the induction hypothesis by $(1+x)$ does not flip the inequal …

• Multiplying the hypothesis $(1+x)^k\ge1+kx$ by $(1+x)$ without first checking $1+x>0$ — if $1+x$ were negative the inequality would reverse. This is exactly why the problem restricts $x>-1$. …