This is my favorite inequality, just because it has such a beautiful proof. Show that, given four positive numbers a, b, c, d such that a+b+c+d=4, that the following holds: \[\fracab+\fracbc+\fraccd+\fracda \le \frac{4}{abcd}\]

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This is my favorite inequality, just because it has such a beautiful proof. Show that, given four positive numbers a, b, c, d such that a+b+c+d=4, that the following holds: \[\fracab+\fracbc+\fraccd+\fracda \le \frac{4}{abcd}\]

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My bad, too many LaTeX shortcuts: \[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\le \frac{4}{abcd}\]
\[ LHS = \frac{a^2cd + b^2ad + c^2ab + d^2bc}{abcd} \] Now ... obviously we want to show somehow that \[ a^2cd + b^2ad + c^2ab + d^2bc \leq 4 \] tick tock. Need a moment on this one.
How about trying to show this? \[a^2cd+b^2ad+c^2ab+d^2bc-a-b-c-d \le 0\]

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Other answers:

\[a^{2}cd + b^{2}ad+c^{2}ab+d^{2}bc \le a + b + c + d\]
Is there an algebraic trick involved? Because I keep hitting walls. :D Very frustrated right now.
\[4(a^{2}cd + b^{2}ad+c^{2}ab+d^{2} bc) \le (a + b + c + d)^2\]\[4(a^{2}cd + b^{2}ad+c^{2}ab+d^{2}bc) \le a^2 + 2ab + 2ac +2ad + b^2 +2bc + 2bd + c^2 + 2cd + d^2\] That still goes nowhere...
I have a proof using Lagrange multipliers, but that is certainly not the 'beautiful proof'. With these sorts of questions there are two standard tools to use - geometric mean < arithmetic mean - Cauchy-Schwartz. For example, here's a result you can obtain using the fact that the geometric mean < arithmetic mean: a+b+c+d=4, then the arithmetic mean of a,b,c,d is equal to 1. Hence \[ \sqrt[4]{abcd} \leq 1 \implies abcd \leq 1 \] A use of Cauchy-Schwartz would be \[ (a/b + b/c + c/d + d/a)^2 = [(a,b,c,d) \cdot (1/b,1/c,1/d,1/a)]^2 \] \[ \leq (a^2 + b^2 + c^2 + d^2)(1/a^2 + 1/b^2 + 1/c^2 + 1/d^2) \] However, both of those results don't get me any closer to proving the result the question asks. Although I'll bet a nickel that some version of them is part of the answer. I'm sure when I see the 'beautiful proof' I'll be angry with myself!

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