Integral Calculus inequalities problem

tomelwood
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Homework Statement


Hey, just wondering how I might go about doing this problem, as I really have very little idea...

Prove the following inequality:
\frac{1}{e}\leq\frac{1}{4\pi^{2}}\int_{R}e^{cos(x-y)}dxdy\leqe
(hopefully this reads "one over e is less than or equal to one over four pi squared times the integral over R of e to the power of cos(x-y) dx dy which is less than or equal to e"

Homework Equations



R is the region [0,2pi]x[0,2pi]

The Attempt at a Solution


I think the Mean value, and intermediate value theorem may come into it somewhere, but I really don't know where to begin. Any pointers at all would be greatly appreciated.
Many thanks
 
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Hint: what is the minimal value of e^{\cos(x-y)} on the whole R?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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