An inequality
in [Math]
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From: William Elliot on 9 Jun 2010 06:05 On Wed, 9 Jun 2010, Henry wrote: > On 9 June, 05:35, William Elliot <ma...(a)rdrop.remove.com> wrote: >> On Tue, 8 Jun 2010, Jim Ferry wrote: >>> Let e(x)=x^x. �Show >>> e(a+b)^2 e(b+c)^2 <= e(a) e(b)^2 e(c) e(a+2b+c) >>> for all a,b,c >= 0, with equality iff b^2 = a c. >> >> Let f(x) = x^x, for all x in [0,oo). >> What's f(0)? �f(0) = lim(x->0) x^x = 0 ? > > lim(x->0) x^x = 1 when the limit is taken from above. > E.g. 0.001^0.001 = 0.993... > 0.0000001^0.0000001 = 0.999998... > You are correct since lim(x->0) x.log x = 0 and e^0 = 1. ----
From: Jim Ferry on 9 Jun 2010 11:03 On Jun 9, 12:35 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Tue, 8 Jun 2010, Jim Ferry wrote: > > Let e(x)=x^x. Show > > e(a+b)^2 e(b+c)^2 <= e(a) e(b)^2 e(c) e(a+2b+c) > > for all a,b,c >= 0, with equality iff b^2 = a c. > > Let f(x) = x^x, for all x in [0,oo). > What's f(0)? f(0) = lim(x->0) x^x = 0 ? > > Show > f(a+b)^2 f(b+c)^2 <= f(a) f(b)^2 f(c) f(a+2b+c) > for all a,b,c >= 0, with equality iff b^2 = a c. > > That statement is false for b = 0. For all a,c >= 0, > 0 <= f(a)^2 f(c)^2 <= f(a) f(0)^2 f(c) f(a + c) = 0 > f(a).f(c) = 0 Yes, let's ditch the e(x) in favor of f(x). I intended f(0) = 1, though I did not state this. For b = 0 the statement follows from a^a <= (a+c)^a and c^c <= (a+c)^c, each of which attains equality iff a*c = 0.
From: Jim Ferry on 11 Jun 2010 09:18
On Jun 9, 11:03 am, Jim Ferry <corkleb...(a)hotmail.com> wrote: > On Jun 9, 12:35 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > > > On Tue, 8 Jun 2010, Jim Ferry wrote: > > > Let e(x)=x^x. Show > > > e(a+b)^2 e(b+c)^2 <= e(a) e(b)^2 e(c) e(a+2b+c) > > > for all a,b,c >= 0, with equality iff b^2 = a c. > > > Let f(x) = x^x, for all x in [0,oo). > > What's f(0)? f(0) = lim(x->0) x^x = 0 ? > > > Show > > f(a+b)^2 f(b+c)^2 <= f(a) f(b)^2 f(c) f(a+2b+c) > > for all a,b,c >= 0, with equality iff b^2 = a c. > > > That statement is false for b = 0. For all a,c >= 0, > > 0 <= f(a)^2 f(c)^2 <= f(a) f(0)^2 f(c) f(a + c) = 0 > > f(a).f(c) = 0 > > Yes, let's ditch the e(x) in favor of f(x). I intended f(0) = 1, > though I did not state this. For b = 0 the statement follows from a^a > <= (a+c)^a and c^c <= (a+c)^c, each of which attains equality iff a*c > = 0. I looked at this a little more, and there's a straightforward, if unilluminating proof. Just take logs of both sides and subtract to get an inequality of the form X >= 0. Take the derivative of X with respect to b: its sign equals that of b^2 - a c, so the minimum is at b = sqrt(a c). At this minimum one can verify that X = 0. |