Bernoulli Inequality

Bernoulli's inequality states: if $x\geq -1$ ${\ displaystyle x \ geq -1}$ $x \ geq -1$ then

(1+x)^{n}\geq 1+nx

{\ displaystyle (1 + x) ^ {n} \ geq 1 + nx}

(1 + x) ^ n \ geq 1 + nx

for all

n\in \mathbb {N} _{0}.

{\ displaystyle n \ in \ mathbb {N} _ {0}.}

n \ in \ mathbb {N} _0.

Proof

The proof of the inequality is carried out by the method of mathematical induction on n . For n = 1, the inequality is obviously true. Suppose that it is true for n , we prove its validity for n +1:

(1+x)^{n+1}=(1+x)(1+x)^{n}\geq (1+x)(1+nx)\geq (1+nx)+x=1+(n+1)x

{\ displaystyle (1 + x) ^ {n + 1} = (1 + x) (1 + x) ^ {n} \ geq (1 + x) (1 + nx) \ geq (1 + nx) + x = 1 + (n + 1) x}

,

t.d.

Generalized Bernoulli inequality

The generalized Bernoulli inequality claims that with $x>-1\!\$ ${\ displaystyle x> -1 \! \}$ and $n\in \mathbb {R}$ ${\ displaystyle n \ in \ mathbb {R}}$ :

if a $n\in (-\infty ;0)\cup (1;+\infty )$ ${\ displaystyle n \ in (- \ infty; 0) \ cup (1; + \ infty)}$ then $(1+x)^{n}\geq 1+nx$ ${\ displaystyle (1 + x) ^ {n} \ geq 1 + nx}$
if a $n\in (0;1)\!\$ ${\ displaystyle n \ in (0; 1) \! \}$ then $(1+x)^{n}\leq 1+nx$ ${\ displaystyle (1 + x) ^ {n} \ leq 1 + nx}$
while equality is achieved in two cases: $\left[{\begin{matrix}\forall x\neq -1,n=0,n=1\\\forall n\neq 0,x=0\end{matrix}}\right.$ ${\ displaystyle \ left [{\ begin {matrix} \ forall x \ neq -1, n = 0, n = 1 \\\ forall n \ neq 0, x = 0 \ end {matrix}} \ right.}$

Evidence

Consider $f(x)=(1+x)^{n}-nx\!\$ ${\ displaystyle f (x) = (1 + x) ^ {n} -nx \! \}$ , and $x>-1,n\neq 0,n\neq 1\!\$ ${\ displaystyle x> -1, n \ neq 0, n \ neq 1 \! \}$ .
Derivative $f'(x)=n(1+x)^{n-1}-n=0\!\$ ${\ displaystyle f '(x) = n (1 + x) ^ {n-1} -n = 0 \! \}$ at $x=x_{0}=0\!\$ ${\ displaystyle x = x_ {0} = 0 \! \}$ , insofar as $n\neq 0\!\$ ${\ displaystyle n \ neq 0 \! \}$ .
Function $f\!\$ ${\ displaystyle f \! \}$ twice differentiable in a punctured neighborhood of a point $x_{0}\!\$ ${\ displaystyle x_ {0} \! \}$ . therefore $f''(x)=n(n-1)(1+x)^{n-2}\!\$ ${\ displaystyle f '' (x) = n (n-1) (1 + x) ^ {n-2} \! \}$ . We get:

$f''(x)>0\!\$ ${\ displaystyle f '' (x)> 0 \! \}$ ⇒ $f(x)\geq f(x_{0})\!\$ ${\ displaystyle f (x) \ geq f (x_ {0}) \! \}$ at $n\in (-\infty ;0)\cup (1;+\infty )$ ${\ displaystyle n \ in (- \ infty; 0) \ cup (1; + \ infty)}$
$f''(x)<0\!\$ ${\ displaystyle f '' (x) <0 \! \}$ ⇒ $f(x)\leq f(x_{0})\!\$ ${\ displaystyle f (x) \ leq f (x_ {0}) \! \}$ at $n\in (0;1)\!\$ ${\ displaystyle n \ in (0; 1) \! \}$

Function value $f(x_{0})=1\!\$ ${\ displaystyle f (x_ {0}) = 1 \! \}$ therefore, the following statements are true:

if a $n\in (-\infty ;0)\cup [1;+\infty )$ ${\ displaystyle n \ in (- \ infty; 0) \ cup [1; + \ infty)}$ then $(1+x)^{n}\geq 1+nx$ ${\ displaystyle (1 + x) ^ {n} \ geq 1 + nx}$
if a $n\in (0;1]\!\$ ${\ displaystyle n \ in (0; 1] \! \}$ then $(1+x)^{n}\leq 1+nx$ ${\ displaystyle (1 + x) ^ {n} \ leq 1 + nx}$

It is easy to see that with the corresponding values $x_{0}=0\!\$ ${\ displaystyle x_ {0} = 0 \! \}$ or $n=0,n=1\!\$ ${\ displaystyle n = 0, n = 1 \! \}$ function $f(x)=f(x_{0})\!\$ ${\ displaystyle f (x) = f (x_ {0}) \! \}$ . Moreover, in the final inequality, the restrictions on $n\!\$ ${\ displaystyle n \! \}$ defined at the beginning of the proof, since equality is fulfilled for them. ■

Notes

Inequality is also true for $x\geq -2$ ${\ displaystyle x \ geq -2}$ (at $n\in \mathbb {N} _{0}$ ${\ displaystyle n \ in \ mathbb {N} _ {0}}$ ) Proof for the case $x\in \left[-2,-1\right)$ ${\ displaystyle x \ in \ left [-2, -1 \ right)}$ can also be carried out by mathematical induction.

$(1+x)^{n+1}+(1+x)^{n}=(1+x)^{n}(1+x+1)\geq (1+nx)(1+x+1)=1+(n+1)x+1+nx(1+x)$ ${\ displaystyle (1 + x) ^ {n + 1} + (1 + x) ^ {n} = (1 + x) ^ {n} (1 + x + 1) \ geq (1 + nx) (1 + x + 1) = 1 + (n + 1) x + 1 + nx (1 + x)}$

Since when $x\in \left[-2,-1\right)$ ${\ displaystyle x \ in \ left [-2, -1 \ right)}$ $(1+x)^{n}\leq 1\leq 1+nx(1+x)$ ${\ displaystyle (1 + x) ^ {n} \ leq 1 \ leq 1 + nx (1 + x)}$ then $(1+x)^{n+1}\geq 1+(n+1)x$ ${\ displaystyle (1 + x) ^ {n + 1} \ geq 1+ (n + 1) x}$

Bernoulli Inequality

Proof

Generalized Bernoulli inequality

Notes

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