Induction principle: $(1+x)^{n+1}>1+(n+1)x$

Question

Given $(1+x)^{n+1}>1+(n+1)x$ for $x>0$ and for $n$ positive integer.

Putting $n=1$ I get $x^2>0$ so the inductive hypothesis is true.

Putting $n=n+1$ I get $(1+x)^{n+2}>1+(n+2)x$. I can write $(1+x)^{n+2}$ as $(1+x)^n(1+x)^2$ and then I don't know how to go on.

Answer 1

No, write

$$(x+1)^{n+2}=(x+1)^{n+1}(x+1)$$ instead of $$(x+1)^{n+2}=(x+1)^n(x+1)^2.$$

Then you can use the induction hypothesis and

$$(x+1)^{n+1}(x+1)>(1+(n+1)x)(1+x)=1+(n+2)x+x^2>1+(n+2)x.$$

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In fact it is easier to start with the equivalent statement

$$(x+1)^n>1+nx$$ for $n>1$.

Then

$$(x+1)^{n+1}>(1+nx)(x+1)>1+(n+1)x$$

Answer 2

Hint

The best way is write

$$(1+x)^{n+2}=x^{n+1}(x+1)>[1+(n+1)x](x+1)$$

Remmember, you have to get $(1+x)^{n+2}>1+(n+2)x$

Can you finish?

Answer 3

Hint: Multiplying your inequality by $1+x>0$ we get $$(1+x)^{n+2}>(1+(n+1)x)(1+x)>1+(n+2)x$$ Show this. The right hand side is: $$1+nx+x+x+nx^2+x^2>1+nx+2x$$ and this is $$x^2(n+1)>0$$

Answer 4

We assume that $(1+x)^{n+1}>1+(n+1)x$

We need to prove that $(1+x)^{n+2}>1+(n+2)x$ is true.

From the first line, we have

$(1+x)(1+x)^{n+1}>(1+x)(1+(n+1)x)=1+(n+1)x+x+x^2>1+(n+2)x$

Answer 5

$\textbf{Proof}$

Since $(1+x)^{2}=1+2x+x^2>1+2x$, the inequality holds for $n=1$. Suppose that it holds for $n=k$, then $$\begin{align*}(1+x)^{k+2}&=(1+x)^{k+1}(1+x)\\&>[1+(k+1)x] (1+x)\\&=1+(k+2)x+(k+1)x^2\\&>1+(k+2)x .\end{align*}$$ This shows that the inequality holds for $n=k+1.$ By mathematical induction, it necessarily holds for all $n \in \mathbb{N_+}.$