TITLE: Proving a multivariate inequality over $0<x<1$ and $n>2$
QUESTION [5 upvotes]: EDIT:  I meant to have the coefficients reversed, showing: 
$$\frac{n}{n-1}(1-(1-x)^n)^n + (1-x)^{n-1} \leq 1$$ 
This version should be true.. but still trying to prove it...
ORIGINAL:
Is it possible to show:
$$(1-(1-x)^n)^n + \frac{n}{n-1}(1-x)^{n-1} \leq 1$$ for $0<x<1$ and  $n\geq 2$ (and $n$ is an integer)? 
This increases the difficulty over the other questions I just asked.  
the second term seems to decrease with $n$, so I can solve for the minimum of $n$.  But the first term doesn't completely increase with $n$ -- the lines cross when I plot the first term with $n=2$ and then $n=3$.  So I'm not sure where to go from there.

REPLY [4 votes]: It fails already for $n=2$. Make the substitution suggested by Thijs, and the left-hand side becomes $(1-y^2)^2+2y = 1+y^4 +2y(1-y) > 1$ for $0<y<1$.

REPLY [3 votes]: This might be false for every positive integer $n \ge 2$.
I believe we can show this for $\displaystyle x \gt 1 - \frac{1}{n-1}$ by using Bernoulli's inequality on the first term on the left side.
Using Bernoulli's.
$$(1 - (1-x)^n)^n \ge 1 - n(1-x)^n$$
And so
$$(1 - (1-x)^n)^n  + \frac{n(1-x)^{n-1}}{n-1} \ge 1 - n(1-x)^n + \frac{n(1-x)^{n-1}}{n-1}$$
$$= 1 - n(1-x)^{n-1}\left((1-x) - \frac{1}{n-1}\right) $$
if
$$ 1 - x \lt \frac{1}{n-1}$$
then 
$$1 - n(1-x)^{n-1}\left((1-x) - \frac{1}{n-1}\right) \gt 1$$
Are you hoping for this to be true, or do you know this to be true (like assigned textbook problem).