Algebraic Proof Using Induction [duplicate]

Let $n$ be a positive integer, and let $x \ge -1$. Prove, using induction, that $$(1 + x)^n \ge 1 + nx.$$

I don't know what to do, I can't expand the left side. I'm not familiar with induction, so can someone please provide an answer? Thanks!

• What's hard in expanding the left side? $(1+x)^{n+1}=(1+x)(1+x)^n$. – Henrik supports the community Aug 12 '16 at 19:45
• You dont want to fully expand the left side. As @Henrik pointed out, you want to expand the $n+1$ case to resemble the $n$ case. – SquirtleSquad Aug 12 '16 at 19:45
• possible dublicate math.stackexchange.com/questions/181702/… – haqnatural Aug 12 '16 at 19:52
• At the end of this answer, is a proof by induction of this inequality, also known as Bernoulli's Inequality. This inequality can be extended to rational exponents by induction as shown at the end of this answer. – robjohn Aug 12 '16 at 20:19
• It's a very useful result. It allows us to show that every positive number has a an n-th root and thus exponents can be expanded to rational and real exponents, and that given any positive b and c there is an x so that $b^x = c$ and thus logarithms are well defined. – fleablood Aug 12 '16 at 21:42

Induction always has two steps.

1) Basic step. Show it is true for $n = 1$. (Although sometime for $n=0$ or another number--- but its always a base case. As we want to prove it for all positive integers we want to start by proving it is true for $n =$ the smallest positive integer).

If $n = 1$ then

$(1 + x)^n = (1 + x)^1 = 1 + x = 1 + 1*x = 1 + nx \ge 1 + nx$.

So it is true for $n = 1$.

2) Inductive step. Show that if is is true for some $n = k$. Then it is true also for $n = k + 1$.

[If we can show this then we know since it is true for $n=k =1$ it will also have to be true for $n = 2; n= 3; n =4;.... n = j; n = j+1; n = j+ 2......$ and therefore be true for all positive integers.]

So if $(1 + x)^k \ge 1 + kx$ then...

$(1 + x)^{k+1} = (1+x)(1+x)^k$.

Now we know $(1 + x)^k \ge 1 + kx$

And as $x \ge -1$ we know $1 + x \ge 0$

So $(1+x)(1+x)^k \ge (1+x)(1 + kx)$

$(1+x)(1+kx) = 1 + x + kx + kx^2$

$= 1 + (k+1)x + kx^2 \ge 1 + (k+1)x$.

So $(1+x)^{k+1} \ge 1+ (k+1)x$

So we have proven that if it is true for $n = k$ it is true for $n = k + 1$.

....

Therefore we are done. (Because it is true for $n = 0$ so it is true for $n =1$. It is true for $n = 1$ so it is true for $n = 2$. It is true for $n = 2$ so it is true for $n = 3$. It is true for.....)

• Thanks for answer, fleablood! – user359548 Aug 12 '16 at 20:25

Hint: $(1+x)^{n+1}=(1+x)^n\cdot(1+x)\geq(1+nx)(1+x)=1+(n+1)x+nx^2$

I know you asked for an inductive proof. Nonetheless, here is a non-inductive proof, using AM-GM. Suppose that $1+nx\geq 0$ (the case $1+nx<0$ being trivial). Then, by AM-GM, $$1+x=\frac{(n-1)\cdot 1+1\cdot (1+nx)}{(n-1)+1}\geq \left(1^{n-1}(1+nx)^1\right)^{\frac{1}{(n-1)+1}}=(1+nx)^{\frac{1}{n}}\,.$$ The result follows immediately (and the equality case is easily seen to be $x=0$ and $n=1$). Note that $n$ can be any real number greater than or equal to $1$.