2
$\begingroup$

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!!! 
$\endgroup$
4
  • $\begingroup$ 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 $(1+x)^k$. $\endgroup$
    – frabala
    Feb 10, 2014 at 2:05
  • $\begingroup$ how to access math mode, how can I write, I am new $\endgroup$
    – hacikho
    Feb 10, 2014 at 2:06
  • $\begingroup$ Here: math.stackexchange.com/editing-help#latex . You use the dollar signs. $\endgroup$
    – frabala
    Feb 10, 2014 at 2:09
  • $\begingroup$ thank for showing me how to edit my question $\endgroup$
    – hacikho
    Feb 10, 2014 at 2:25

2 Answers 2

2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Does it extend to irrationals? $\endgroup$
    – Emily
    Feb 10, 2014 at 3:57
  • $\begingroup$ @Arkamis: By continuity from the rational case, it does. $\endgroup$
    – robjohn
    Feb 10, 2014 at 4:17
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.