c/(1+c) < a/(1+a) + b/(1+b)
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From: Sam Stump on 29 Mar 2010 13:37 Let a,b,c > 0 and c < a + b, show that c/(1+c) < a/(1+a) + b/(1+b). Any hints? The chapter text leads me to believe that the triangle inquality should be involved, but I just can't see it.
From: Robert Israel on 29 Mar 2010 15:01 > Let a,b,c > 0 and c < a + b, show that c/(1+c) < a/(1+a) + b/(1+b). > > Any hints? > The chapter text leads me to believe that the triangle inquality > should be involved, but I just can't see it. Hint: What t > 0 would make t/(1+c) = a/(1+a) + b/(1+b)? If the inequality were false, you'd have t <= a+b. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: W^3 on 29 Mar 2010 20:23 In article <95d162e6-eb71-4a98-bb5f-10dddc6388f9(a)k19g2000yqn.googlegroups.com>, Sam Stump <sstump64(a)gmail.com> wrote: > Let a,b,c > 0 and c < a + b, show that c/(1+c) < a/(1+a) + b/(1+b). The function x -> x/(1+x) is strictly increasing on (0, oo), so c/(1+c) < (a+b)/(1+(a+b)) = a/(1+(a+b)) + b/(1+(a+b)) < a/(1+a) + b/(1+b).
From: Zdislav V. Kovarik on 31 Mar 2010 14:22 On Mon, 29 Mar 2010, Sam Stump wrote: > Let a,b,c > 0 and c < a + b, show that c/(1+c) < a/(1+a) + b/(1+b). > > Any hints? > The chapter text leads me to believe that the triangle inquality > should be involved, but I just can't see it. Plain high-school arithmetic. Simplify and factor a/(1+a) + b/(1+b) - (a+b)/(1+a+b) If it is positive, you're done because x/(1+x) is increasing for x>0. Cheers, ZVK(Slavek).
From: Ronald Bruck on 31 Mar 2010 16:23
In article <Pine.WNT.4.58.1003311319540.-236349(a)satori.mcmaster.ca>, Zdislav V. Kovarik <kovarik(a)mcmaster.ca> wrote: > On Mon, 29 Mar 2010, Sam Stump wrote: > > > Let a,b,c > 0 and c < a + b, show that c/(1+c) < a/(1+a) + b/(1+b). > > > > Any hints? > > The chapter text leads me to believe that the triangle inquality > > should be involved, but I just can't see it. > > Plain high-school arithmetic. > Simplify and factor > a/(1+a) + b/(1+b) - (a+b)/(1+a+b) > If it is positive, you're done because x/(1+x) is increasing for x>0. One way, surely. Another way: let f : [0,oo) --> [0,oo) satisfy f(0) = 0, f is increasing, and t --> f(t)/t is decreasing (all of which are satisfied by f(t) = t/(1+t). Then f(a+b) c < a + b ==> f(c) < f(a+b) = (a+b) ------ a+b f(a+b) f(a+b) = a ----- + b ------ a+b a+b f(a) f(b) <= a ---- + b ---- a b = f(a) + f(b). Any concave, increasing function with f(0) = 0 suffices. -- Ron Bruck |