# How to prove that $(1 - x)^y \leq 1$, When $1 > x > 0 \wedge y \geq 1$?

The statement in the question is obviously true. But when I tried

FullSimplify[(1 - x)^y <= 1, 1 > x > 0 && y >= 1]


I got

(*(1 - x)^y <= 1*)


Is there any way to get this done through Mathematica?

• Taking logs on both sides seems to help: In:= FullSimplify[y*Log[(1 - x)] <= 0, 1 > x > 0 && y >= 1] Out= True Nov 15, 2018 at 16:31

Try this:

Reduce[(1 - x)^y > 1 && 1 > x > 0 && y >= 1]

(* False  *)


Have fun!

• That is rather weird. How come Reduce[(1 - x)^y <= 1 && 1 > x > 0 && y >= 1] does not work? Nov 15, 2018 at 15:32
• Come on, if one statement is false, the opposite one is true. What is your aim: to solve a problem, or to do it in one special way? Nov 15, 2018 at 16:18
• @ablmf In what sense did your variant fail to work? Nov 15, 2018 at 16:29
• @DanielLichtblau It does not return True. Nov 15, 2018 at 16:59
• @ablmf: Reduce doesn't return a True or False value unless the listed statement is automatically true or automatically False. In your version, the first clause (1 - x)^y <= 1 does follow automatically from the other two, which means if you run Reduce[(1 - x)^y <= 1 && 1 > x > 0 && y >= 1, Reals], you get back 1 > x > 0 && y >= 1. But there are values of $x$ and $y$ for which this latter statement isn't true, so it can't be reduced further. Nov 15, 2018 at 18:10

      Reduce[(1 - x)^y <= 1 && 1 > x > 0 && y >= 1, {x, y}, Reals]