Re: analysis

Posted by Amit Goyal on Oct 29, 2011; 2:32pm
URL: http://discussion-forum.276.s1.nabble.com/analysis-tp6943219p6943417.html

I am going to show:
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ Max{x(k) +y(k): 1 ≤ k ≤ n}

By definition of Max(),
Max{x(k):1 ≤ k ≤ n} ≥ x(j) for all 1 ≤ j ≤ n  ---(1)
Max{y(k):1 ≤ k ≤ n} ≥ y(j) for all 1 ≤ j ≤ n  ---(2)
Adding (1) and (2), we get
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ x(j) +y(j) for all 1 ≤ j ≤ n
This implies that:
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ Max{x(j) +y(j): 1 ≤ j ≤ n} or equivalently, if you prefer,
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ Max{x(k) +y(k): 1 ≤ k ≤ n}