If $p + q = 1$ prove that for any natural $n, m$ following is true: $(1 - p^n)^m + (1 - q^m)^n \ge 1$ Let $p, q \in \mathbb R$ be positive reals for which $p + q = 1$. How to prove that for any two natural numbers $n, m$ the following inequality is true?
$(1 - p^n)^m + (1 - q^m)^n \ge 1$
I don't have a big knowledge about solving inequalities, so I tried to use cauchy-schwarz inequality, binomial theorem and some other baisc techniques, but it lead me nowhere. I've been thinking about it for a long time and now I'm completly stuck.

Let there  are two coins each with probability of head $p$. Let coin 1 has been tossed $n$ times independently and this whole scheme is repeated independently for $m$ times. 
Similarly coin 2 is tossed $m$ times independently and this whole scheme is repeated independently for $n$ times and independently of coin 1.
$A$= at least one tail occurs in the string of 1st $n$-tosses, 2nd $n$ tosses, $\cdots$, $m^{th}$ $n$-tosses for coin 1
$B$= at least one head occurs in the string of 1st $m$-tosses, 2nd $m$ tosses, $\cdots$, $n^{th}$ $m$-tosses for coin 1
Notice the desired form is $P(A\cup B)\leq1$
