## Solution

The problem of this contest was to proof an equivalence:

I also just noticed that this inequation only holds for uneven, positive values of n(which is probably the reason why I had no entries this week).

First of all a small explanation why this is even true:

if x is bigger than 0: Then 1/x^n is only < 1 if x > 1.

if x is smaller 0: Then 1/x^n is only < -1 if x > -1

These two cases connect both inequations.

In this case you can rearrange the leftern inequation to:

`1 < x^(n+1) if x > 0`

→ `x > 1`

`1 > x^(n+1) if x < 0`

→ `1 > ±x`

→ `-1 < ±x`

→ `-1 < x`

The ± comes from taking an even root.

So if x > 0 than x > 1 is the only solution and if x < 0, x > -1 is the only solution.

Putting the 2 together gets the rightern side of the equivalence:

`x > 1`

or `-1 < x < 0`

And the way back to prove it actually is an equivalence:

`x > 1`

→ `x^(n+1) > 1`

→ `x > 1/x^n`

`-1 < x < 0`

→ `1 < x^(n+1)`

→ `x > 1/x^n`

*↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓*

### List of participants with their entries:

Name | solution found | comment |
---|---|---|

@tonimontana | none | found one of the errors I made → you get the reward |

*↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓*

## Winner draw:

Not necessary since only 1 participant:

Congratulations @tonimontana , you won 2 SBI!

*↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓*