Question

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: \[\fracab+\fracbc+\fraccd+\fracda \le \frac{4}{abcd}\]

Answer 1

My bad, too many LaTeX shortcuts:

\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\le \frac{4}{abcd}\]

Answer 2

\[ LHS = \frac{a^2cd + b^2ad + c^2ab + d^2bc}{abcd} \]
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.

Answer 3

How about trying to show this?
\[a^2cd+b^2ad+c^2ab+d^2bc-a-b-c-d \le 0\]

Answer 4

\[a^{2}cd + b^{2}ad+c^{2}ab+d^{2}bc \le a + b + c + d\]

Answer 5

Is there an algebraic trick involved? Because I keep hitting walls. :D Very frustrated right now.

Answer 6

\[4(a^{2}cd + b^{2}ad+c^{2}ab+d^{2} bc) \le (a + b + c + d)^2\]\[4(a^{2}cd + b^{2}ad+c^{2}ab+d^{2}bc) \le a^2 + 2ab + 2ac +2ad + b^2 +2bc + 2bd + c^2 + 2cd + d^2\] That still goes nowhere...

Answer 7

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!