Editing Bernoulli's inequality (section)

==Alternative proofs==
===Arithmetic and geometric means===
An elementary proof for <math>0\le r\le 1</math> and <math>x \ge -1</math> can be given using [[Inequality of arithmetic and geometric means#Weighted AM–GM inequality|weighted AM-GM]].

Let <math>\lambda_1, \lambda_2</math> be two non-negative real constants. By weighted AM-GM on <math>1,1+x</math> with weights <math>\lambda_1, \lambda_2</math> respectively, we get

:<math>\dfrac{\lambda_1\cdot 1 + \lambda_2\cdot (1+x)}{\lambda_1+\lambda_2}\ge \sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}}.</math>

Note that

:<math>\dfrac{\lambda_1\cdot 1 + \lambda_2\cdot (1+x)}{\lambda_1+\lambda_2}=\dfrac{\lambda_1+\lambda_2+\lambda_2x}{\lambda_1+\lambda_2}=1+\dfrac{\lambda_2}{\lambda_1+\lambda_2}x</math>

and

:<math>\sqrt[\lambda_1+\lambda_2]{(1+x)^{\lambda_2}} = (1+x)^{\frac{\lambda_2}{\lambda_1+\lambda_2}},</math>

so our inequality is equivalent to

:<math>1 + \dfrac{\lambda_2}{\lambda_1+\lambda_2}x \ge (1+x)^{\frac{\lambda_2}{\lambda_1+\lambda_2}}.</math>

After substituting <math>r = \dfrac{\lambda_2}{\lambda_1+\lambda_2}</math> (bearing in mind that this implies <math>0\le r\le 1</math>) our inequality turns into

:<math>1+rx \ge (1+x)^r</math>

which is Bernoulli's inequality for <math>0\le r\le 1</math>.

The case <math>r\ge 1</math> can be derived from <math>0\le r\le 1</math> in the same way as 
the case <math>0\le r\le 1</math> can be derived from <math>r\ge 1</math>, see above "Generalization of exponent".

===Geometric series===
Bernoulli's inequality

{{NumBlk|:|<math>(1+x)^r \ge 1+rx </math>|1}}

is equivalent to

{{NumBlk|:|<math>(1+x)^r - 1-rx \ge 0,</math>|2}}
and by the formula for [[geometric series]] (using ''y'' = 1&thinsp;+&thinsp;''x'') we get
{{NumBlk|:|<math> (1+x)^r - 1 = y^r-1 = \left(\sum^{r-1}_{k=0}y^k\right) \cdot (y-1) = \left(\sum^{r-1}_{k=0}(1+x)^k\right)\cdot x</math>|3}}  
which leads to
{{NumBlk|:|<math>(1+x)^r - 1-rx = \left(\left(\sum^{r-1}_{k=0}(1+x)^k\right) - r\right)\cdot x = \left(\sum^{r-1}_{k=0}\left((1+x)^k-1\right)\right)\cdot x \ge 0.</math>|{{EquationRef|4}}}}

Now if <math>x \ge 0</math> then by monotony of the powers each summand <math>(1+x)^k - 1 = (1+x)^k - 1^k \ge 0</math>, and therefore their sum is greater <math>0</math> and hence the product on the [[Sides of an equation|LHS]] of ({{EquationNote|4}}).

If <math> 0 \ge x\ge -2 </math> then by the same arguments <math>1\ge(1+x)^k</math> and thus
all addends <math>(1+x)^k-1</math> are non-positive and hence so is their sum. Since the product of two non-positive numbers is non-negative, we get again
({{EquationNote|4}}).

===Binomial theorem===
One can prove Bernoulli's inequality for ''x'' ≥ 0 using the [[binomial theorem]]. It is true trivially for ''r'' = 0, so suppose ''r'' is a positive integer. Then <math>(1+x)^r = 1 + rx + \tbinom r2 x^2 + ... + \tbinom rr x^r.</math> Clearly <math>\tbinom r2 x^2 + ... + \tbinom rr x^r \ge 0,</math> and hence <math>(1+x)^r \ge 1+rx</math> as required.

===Using convexity===
For <math>0\neq x> -1</math> the function <math>h(\alpha)=(1+x)^\alpha</math> is strictly convex. Therefore, for <math>0<\alpha<1</math> holds
<math>(1+x)^\alpha=h(\alpha)=h((1-\alpha)\cdot 0+\alpha\cdot 1)<(1-\alpha) h(0)+\alpha h(1)=1+\alpha x</math>
and the reversed inequality is valid for <math>\alpha<0</math> and <math>\alpha>1</math>.

Another way of using convexity is to re-cast the desired inequality to  <math>\log (1 + x) \geq \frac{1}{r}\log( 1 + rx)</math> for real <math>r\geq 1</math> and real  <math>x > -1/r</math>. This inequality can be proved using  the fact that the <math>\log</math>  function is concave, and  then using Jensen's inequality in the form <math> \log( p \, a + (1-p)b ) \geq p\log(a) + (1-p)\log(b) </math> to give:
<math>\log(1+x) = \log(\frac{1}{r}(1+rx)+\frac{r-1}{r}) \geq 
\frac{1}{r} \log (1+rx)+\frac{r-1}{r}\log 1 = \frac{1}{r} \log (1+rx)
</math> 
which is the desired inequality.