Is My Induction Proof for (1+ny) ≤ (y+1)n Correct?

[nascentmind]

Homework Statement

Using induction I need to prove (1+ny) $$\leq$$ (y+1)ⁿ

Homework Equations

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The Attempt at a Solution

For n = 1. 1+y = y+1.

For some n = k
(1+ky) $$\leq$$ (y+1)^k

Now to prove for k+1
(1+ky+y)$$\leq$$ (y+1)^k+y
Now I have to prove that (y+1)^k+y $$\leq$$ (y+1)^k+1

By simply expanding (y+1)^k (y+1) can we can see it is greater?

Am I right to this point in solving the problem? If not please provide just a hint.

 

[Filip Larsen]

You are not wrong, but you can only prove the last step assuming something about the value of y.

 

[nascentmind]

The real question is actually (1+x)^1/n - 1 $$\leq$$ x/n .

I have substituted x/n = y. That was the hint given in the book. Here x is a real number and n is an integer.
So now y would be a real number.

 

[Filip Larsen]

Let me rephrase: can you prove your inequality if, say, y = -2?

 

[nascentmind]

Yes I can. -3 < 1 for k = 1.

 

[Filip Larsen]

That only proves the base case. Can you prove it for all n > 0?

 

[nascentmind]

You mean for all y>0 instead of n>0?

 

[Filip Larsen]

I mean n > 0.

Perhaps my questions are more confusing than helpful, so let me be more direct. In your induction step you want to show that

(1) $$(1+ky)+y \leq (1+y)^k + y(1+y)^k$$

given the premises that $1+ky \leq (1+y)^k$. This means that if you can prove

(2) $$y \leq y(1+y)^k$$

then you can prove (1), since $a \leq b$ and $c \leq d$ implies $a+c\leq b+d$. To prove (2) you have two cases, y > 0 and y < 0. The later result in $(1+y)^k \leq 1$ which is clearly false for some values of y and k, so that means you can only prove your original inequality by induction when y > 0, and you therefore have to make additional analysis to prove or disprove if your original inequality holds for y < 0.


(I'm off for new year preparations and won't follow this discussion for a while).

 

[nascentmind]

Re-read the problem and it mentions x is a positive real number and n is a positive integer.
Hence y = x/n > 0
Correct me if I am wrong
Now to prove
y $$\leq$$ y(1+y)^k

If y > 0 is true then y + 1 > 1. (Adding 1 to both sides)
Hence (y+1)^k > 1
Multiplying y on both the sides we have y (y+1)^k > y