PROVE if $x \ge-1 $then $ (1+x)^n \ge 1+nx $ , Every $n \ge 1$

Question

Use mathematical induction to prove this. Here is my answer but I stuck at certain point.

Base Case: n=1 $$(1+x)^1 \ge 1+x $$ True ,

Induction Case: n=k assume $$(1+x)^k \ge 1+kx $$ n=k+1 $$ (1+x)^k+1 \ge 1+(k+1)x $$ $$(1+x)^k *(1+x) \ge 1+ kx+ x $$

      Stuck!!!

Please, fix your post. You didn't write what you want to prove. Also, you can use math-mode to write mathematics in your post, like $(1+x)^k$. — frabala, Feb 10 '14 at 02:05
Here: http://math.stackexchange.com/editing-help#latex . You use the dollar signs. — frabala, Feb 10 '14 at 02:09

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

Bernoulli's Inequality is proven for integer exponents in this answer by induction and then expanded to rational exponents in this answer.

edited Apr 13 '17 at 12:21

Community

answered Feb 10 '14 at 02:23

robjohn

Does it extend to irrationals? – Emily Feb 10 '14 at 03:57
@Arkamis: By continuity from the rational case, it does. – robjohn Feb 10 '14 at 04:17

score 1 · Accepted Answer · answered Feb 10 '14 at 02:17

1

This is the Bernoulli Inequality, whose proof (by induction) can be found on Wikipedia.

answered Feb 10 '14 at 02:17

Yiyuan Lee

2 Answers2