My Math Forum Strange sup & inf

 Real Analysis Real Analysis Math Forum

 November 7th, 2017, 04:52 AM #1 Newbie   Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 Strange sup & inf Find the sup and inf of this set $$A={ \frac{mn}{1+m+n}}$$ With $m,n\in \mathbb{N}$. (Let be A a subset of $\mathbb{R}$). How to find them? I tried to change the variables, putting $a=n+m$ or $a=nm$ but it doesn't seem to work! Any idea?
 November 7th, 2017, 04:59 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra Think geometrically. What could $2(m+n)$ represent? And what then is $mn$? For any given value of $m+n$, how can you maximise/minimise $mn$? Last edited by v8archie; November 7th, 2017 at 05:07 AM.
 November 7th, 2017, 01:47 PM #3 Global Moderator   Joined: May 2007 Posts: 6,581 Thanks: 610 There is no sup. For example, let m=n and let it get bigger, so A is approximately n (no bound). For inf, let m=n=1, then A=1/3. You need to verify that for larger values for n or m, A is larger.
 November 10th, 2017, 12:40 AM #4 Newbie   Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 Thanks, I proved it. $$\frac{mn}{1+m+n}\ge \frac{1}{3} \rightarrow 3mn \ge m+n+1 \rightarrow m(n-1)+n(m-1)+mn-1 \ge 0$$ This is correct since $m,n \ge 1$.

 Tags inf, real analysis, sets, strange

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Suliman Calculus 1 September 20th, 2013 11:33 PM capea Real Analysis 0 September 29th, 2011 11:12 AM proglote Algebra 9 August 24th, 2011 03:35 PM fathwad Number Theory 10 May 28th, 2007 01:42 PM bigli Real Analysis 3 May 27th, 2007 03:31 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top