Math, asked by mp643981, 1 month ago

what is the Behrouli inequelity​

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Answered by ankaneha5
0

Answer:

In mathematics, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x. It is often employed in real analysis.

An illustration of Bernoulli's inequality, with the graphs of {\displaystyle y=(1+x)^{r}}{\displaystyle y=(1+x)^{r}} and {\displaystyle y=1+rx}{\displaystyle y=1+rx} shown in red and blue respectively. Here, {\displaystyle r=3.}r=3.

The inequality states that

{\displaystyle (1+x)^{r}\geq 1+rx}{\displaystyle (1+x)^{r}\geq 1+rx}

for every integer r ≥ 0 and every real number x ≥ −1.[1] If the exponent r is even, then the inequality is valid for all real numbers x. The strict version of the inequality reads

{\displaystyle (1+x)^{r}>1+rx}{\displaystyle (1+x)^{r}>1+rx}

for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0.

There is also a generalized version that says for every real number r ≥ 1 and real number x ≥ −1,

{\displaystyle (1+x)^{r}\geq 1+rx,}{\displaystyle (1+x)^{r}\geq 1+rx,}

while for 0 ≤ r ≤ 1 and real number x ≥ −1,

{\displaystyle (1+x)^{r}\leq 1+rx.}{\displaystyle (1+x)^{r}\leq 1+rx.}

Bernoulli's inequality is often used as the crucial step in the proof of other inequalities. It can itself be proved using mathematical induction, as shown below.

Answered by parmarjayendra76
0

Answer:

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Step-by-step explanation:

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