You can prove that it is a strict inequality... The way I found looks a bit messy ><
Prove that there isn't such that (we can exlude 1, because it's not equal when x=1)
But since x is in [0,1), 2x
Now what's the sign of ?
Differentiate and you'll get :
, which is always positive, for any x in [0,1)
Thus is increasing in [0,1)
So and we have
So there exists a unique value such that (by the intermediate value theorem, or something like that)
and for any and for any
But . So they have the same sign.
So , for any
Now you're up to prove that :
We know that is such that , that is to say :
By , we have :
Note that << MISTAKE
We can easily see that , because .
And similarly, we can see that , because
I hope this ... erm helps ^^
Edit : okay, because of the mistake pointed out above, the conclusion is false... I'm too tired to think again about it, sorry ><