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Show that for every real number $x\geqslant -1$ and every positive integer $n$, $(1+x)^n \geqslant 1+nx$.

This is what i have so far

Base case:
    n=1,x=-1
    (1+(-1))^1 >= 1+(1)(-1)
    0 >= 0 This is true

Inductive case:
Assume every real number y>=1 and every positive integer k: 
(1+y)^k >= 1+ky is true.  And I want to show that for k+1 and y+1
(1(y+1))^(k+1) >= 1 + (k+1)(y+1)
(y+2)^(k+1) >= 1 + (k+1)(y+1)
(y+2)^k * y+2 >= 1 + (k+1)(y+1)

I am lost from here. Since we assumed that $(1+y)^k\geqslant 1+ky$ can we say that by induction hypothesis $(y+2)^k \cdot y+2 \geqslant 1+(k+1)(y+1)$ also? Any Help is appreciated. Thanks in advance.

ki3i
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Abhi
  • 237

1 Answers1

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Seems you are wrongly doing the induction on y. It should be done on k.

OK, you have $(1+y)^k \geq 1+ky$ by the induction hypothesis.

Then you can write that:

$(1+y)^{(k+1)} = (1+y)(1+y)^k \geq (1+y)(1+ky) = 1 + (k+1)y + ky^2 \geq 1 + (k+1)y$

The inequalities are true because $y+1\geq0$ and $ky^2\geq0$.
Thus you have it now proved for $k+1$.

This is the famous Bernoulli's inequality.
You could have looked up the proof in a lot of sources.

peter.petrov
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