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!!! 
$\endgroup$
  • $\begingroup$ 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 '14 at 2:05
  • $\begingroup$ how to access math mode, how can I write, I am new $\endgroup$ – hacikho Feb 10 '14 at 2:06
  • $\begingroup$ Here: math.stackexchange.com/editing-help#latex . You use the dollar signs. $\endgroup$ – frabala Feb 10 '14 at 2:09
  • $\begingroup$ thank for showing me how to edit my question $\endgroup$ – hacikho Feb 10 '14 at 2:25
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Does it extend to irrationals? $\endgroup$ – Emily Feb 10 '14 at 3:57
  • $\begingroup$ @Arkamis: By continuity from the rational case, it does. $\endgroup$ – robjohn Feb 10 '14 at 4:17

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.