Proof $x$, $1+nx≤ (1+x)^n$

Question

Prof using the binomial theorem: for all integers $n ≥0$ and for all nonnegative real numbers $x$, $1+nx ≤(1+x)^n$.

Don't have a idea to start this one. I don't know how to use math induction yet, so I need a answer without that.

Answer 1

Using Binomial theorem : $$(1+x)^{n} = \sum_{0}^{n}{{{k}\choose{n}}x^{k}} $$ When consider $k=1,k=0$.

Answer 2

Using the binomial theorem:

$$(1+x)^n=1+nx+\dots$$

Where $\dots$ is non-negative, as it is a polynomial in $x$ with non-negative coefficients and $x\geq 0$.

Answer 3

Hint: $$1=\dbinom{n}{0}1^{n-0}x^0\quad \text{ and } \quad nx=\dbinom{n}{1}1^{n-1}x^1$$

Answer 4

I guess an easy way to do it would be induction if you are having trouble with binomial theorem. For base case n=1 it is trivial $$1+x \le 1+x$$ Let it be true for some $n=k$. We have $$\begin{align} 1+kx \le(1+x)^k\\ \Rightarrow (1+kx)(1+x) \le (1+x)^{k+1}\\ \Rightarrow 1+x(k+1)+kx^2\le(1+x)^{k+1}\\ \end{align}$$ But if $k\gt 0$ and $x\gt0$ we must have $$\Rightarrow 1+x(k+1)+kx^2 \ge 1+(k+1)x$$ Therefore we have $$1+(k+1)x \le (1+x)^{k+1}$$ Hence by induction $$1+nx \le (1+x)^n$$