Inequality of Bernoulli

Definition

The inequality of Bernoulli stipulates that:

$(1+x)\; ^n\; >\; 1+nx~$

for all Whole naturalness n>=2, and any number real X not no one and strictly higher than -1.

Demonstration

Soit$x\; \backslash \; in\; \backslash \; mathbb\; \{R^\; \{\backslash \; star\}\}$ such as $x\; >\; -1~$ and $m\; \backslash \; in\; \backslash \; mathbb\; \{NR\}$ such as $m\; \backslash \; ge2$ and one seeks to show that $\backslash \; left\; (1+x\; \backslash \; right)\; ^m\; >\; 1+mx$

One will define the function $f$ definite on $\backslash \; mathbb\; \{R\}$ by:

$f\; \backslash \; left\; (X\; \backslash \; right)\; =\; \backslash \; left\; (1+x\; \backslash \; right)\; ^m\; -\; \backslash \; left\; (1+mx\; \backslash \; right)$

One will show that the function $f\; >\; 0$ on the interval $\backslash \; left]\; -\; 1,0\; \backslash \; right\; \backslash \; cup\; \backslash \; left]\; 0,\; +\; \backslash \; infty\; \backslash \; right\; The\; derivative\; of\; the\; function\; on\; the\; field\; considered\; is:$

$f\text{'}\; \backslash \; left\; (X\; \backslash \; right)\; =m\; \backslash \; left\; (1+x\; \backslash \; right)\; ^\; \{M-1\}\; -\; m$

$f\text{'}\; \backslash \; left\; (X\; \backslash \; right)\; =m\; \backslash \; left\; (\backslash \; left\; (1+x\; \backslash \; right)\; ^\; \{M-1\}\; -\; 1\; \backslash \; right)$

One now studies the sign of the derived one:

$f\text{'}\; \backslash \; left\; (X\; \backslash \; right)\; =0\; \backslash \; Longleftrightarrow\; x=0$

for $x\; \backslash \; in\; \backslash \; left]\; -\; 1,0\; \backslash \; right\; and\; :$ f\text{'}\; \backslash \; left\; (X\; \backslash \; right)\; >0$for$ X\; \backslash \; in\; \backslash \; left]\; 0,\; +\; \backslash \; infty\; \backslash \; right$$

The function $f$ is thus strictly decreasing on the interval $\backslash \; left]\; -\; 1,0\; \backslash \; right\; and\; strictly\; increasing\; on\; the\; interval$ \backslash \; left0,\; +\; \backslash \; infty\; \backslash \; right\; For$ x=0~$,\; there\; is$ \backslash \; left\; (1+x\; \backslash \; right)\; ^m\; -\; \backslash \; left\; (1+mx\; \backslash \; right)\; =0$$$
There is thus well $f\; >\; 0$ on the interval $\backslash \; left]\; -\; 1,0\; \backslash \; right\; \backslash \; cup\; \backslash \; left]\; 0,\; +\; \backslash \; infty\; \backslash \; right$

Another demonstration

Here a demonstration by recurrence

1) Initialization :

For n=2 by supposing X not no one one a:

$1+2x+x^2>1+2x~$

or:

$(1+x)\; ^2>1+2x~$

Thus the property is true with row 2.

2) Heredity :

Assumption of recurrence: $(1+x)\; ^k>1+kx~$

Let us show that the property is true with the following row k+1:

$(1+x)\; ^k>1+kx~$

$(1+x)\; ^k\; (1+x)\; >\; (1+kx)\; (1+x)\; ~$ indeed one does not change the direction of the inequality because one thus supposes x>-1 x+1>0

$(1+x)\; ^\; \{k+1\}\; >1+x+kx+kx^2~$

However

$1+x+kx+kx^2=1+\; (k+1)\; x+kx^2>1+\; (k+1)\; x~$

From where

$(1+x)\; ^\; \{k+1\}\; >1+\; (k+1)\; x~$

3) Conclusion:

The property is true with row 2 and it is hereditary thus true for entire N equal to or higher than 2 with X not no one and strictly higher than -1.

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