Another short solution: ... ... ... ... ... ... ... ... ... ... Using Jensen's: log[ Integral_{0 ≤ x ≤ 1} f(x+c)/f(x) dx ] >= Integral_{0 ≤ x ≤ 1} log(f(x+c)/f(x)) dx = 0 These solutions are related as Jensen's can be used (again with the log function) to quickly prove the AM/GM inequality. J.P. On Wed, Jul 8, 2020 at 4:51 PM Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 08/07/2020 20:57, Dan Asimov wrote:
Let f : R —> R
be a continuous function that is positive and periodic of period 1. That is,
f(x+1 ) = f(x)
for all x.
Prove that for any real number c
Integral_{0 ≤ x ≤ 1} f(x+c)/f(x) dx ≥ 1
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The integral is clearly a continuous function of c (note: f is bounded below because it's a continuous function on a compact space, namely R/Z). So it's enough to prove the theorem when c=p/q is rational.
Now write g(x) = f(x+c)/f(x) + f(x+2c)/f(x+c) + etc., a sum of q terms. Our integral is equal to 1/q times the integral of g over the unit interval.
But g(x) >= q by the AM-GM inequality, and we're done.
-- g
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