Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.
$P\left(n\right):(1+x{)}^n\ge (1+nx)$, for $x>-1$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):(1+x)\ge (1+x)$ for $x>-1$
Thus, $P\left(n\right)$ is true for $n=1$.

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$.
Put $n=K$ in $P\left(n\right)$, and assume that $P\left(K\right)$ is true for some natural number $K$ i.e.
$P\left(K\right):(1+x{)}^K\ge (1+Kx),x>-1$ is true. $\cdots \left(1\right)$

Step $4:$ Checking statement $P\left(n\right)$ for $n=K+1$.
Now, we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
$(1+x{)}^{K+1}=(1+x{)}^K(1+x)$
Given that $x>-1$, so $(1+x)>0$
Therefore, by using $(1+x{)}^K\ge (1+xK)$, we have
$(1+x{)}^{K+1}\ge (1+Kx\left)\right(1+x)$
$\Rightarrow (1+x{)}^{K+1}\ge (1+x+Kx+K{x}^2)$
Here, $K$ is a natural number and $x^2\ge 0$, so that $K{x}^2\ge 0$.
Therefore,
$(1+x+Kx+K{x}^2)\ge (1+x+Kx)$
So, we obtain
$(1+x{)}^{K+1}\ge (1+x+Kx)$
i.e., $(1+x{)}^{K+1}\ge [1+x(1+K)]$
Thus, $P(K+1)$ is true when $P\left(K\right)$ is true.
Final Answer:
Hence, by the principle of Mathematical Induction, $P\left(n\right)$ is true for all-natural numbers.