Prove that, $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Prove that, $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for a little advanced and elegant solution.  
EDIT: $a,b,c,d>0$
 A: For $N > 0$, consider the set
$$
D_N = \left\{ (a,b,c,d) \in \mathbb R^4 \, | \, \frac 1N \le  a,b,c,d \le N \right\}.
$$
This set is compact, and the function $f(a,b,c,d) \overset{def}= \frac ab + \frac bc + \frac cd + \frac da$ is continuous on $D_N$, hence we have the existence of a minimum. Without loss of generality, we can assume that there is a minimizer in the interior of $D_N$, since if it is on the boundary for some $N$, it is in the interior of a $D_N$ for $N$ slightly bigger. (Note that when any component approaches $0$ or $\infty$, $f \to \infty$, hence the minimum is not there, which allows us to consider the interior only and dismiss boundaries.)
Computing the gradient, one gets 
$$
\nabla f(a,b,c,d) = \left( \frac 1b - \frac d{a^2}, \frac 1c - \frac a{b^2}, \frac 1d - \frac b{c^2}, \frac 1a - \frac c{d^2}\right)
$$
and since our minimizer will be in the interior, there exists a minimum (A,B,C,D) such that 
$$
A^2 = BD, B^2 = CA, C^2 = DB, D^2 = AC \implies A=B=C=D \implies f(A,B,C,D) = 4.
$$
Since $4$ is the minimum value for all $N$, it is the minimal value. 
Hope that helps,
A: Let's prove that $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'} \geq 4$ if $(a', b', c', d')$ is the tuple $(a, b, c, d)$ shuffled in an arbitrary order. Your inequality will follow as a special case when $(a', b', c', d')=(b, c, d, a)$.
When the problem is formulated like this, we can safely assume that $a \leq b \leq c \leq d$. Let's do that to make things easier.
If $(a', b', c', d') \neq (a, b, c, d)$, then there's a pair of adjacent elements in the tuple $(a', b', c', d')$ that goes in decreasing order. For instance, let's assume that $a' > b'$. Form a new tuple $(a'', b'', c'', d'') = (b', a', c', d')$. We have swapped positions of the two elements so that now they go in the ascending order. Observe that since $a' > b'$ and $a < b$, we have
$$
\frac{a}{a''} + \frac{b}{b''} = \frac{a}{b'} + \frac{b}{a'} \leq \frac{a}{a'} + \frac{b}{b'},
$$
therefore
$$
\frac{a}{a''} + \frac{b}{b''} + \frac{c}{c''} + \frac{d}{d''} \leq
\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'}.
$$
Using this observation, we can do the following. We can take the tuple $(a', b', c', d')$ and bubble sort it. Each time when we swap places of two adjacent elements that go in descending order, the value $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'}$ decreases, as we have just shown. When we are done with the bubble sort, we will have the tuple that goes in ascending order, i.e. $(a', b', c', d') = (a, b, c, d)$. For this tuple we have $\frac{a}{a} + \frac{b}{b} + \frac{c}{c} + \frac{d}{d} = 4$. Since the value was only decreasing during the bubble sort, the original value was greater or equal than the final one. Therefore $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'} \geq 4$, for an arbitrary arrangement $(a', b', c', d')$, qed.
A: Not exactly AM-GM, but Rearrangement Inequality immediately gives
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d} = 4$$
A: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)=$$
$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{\left(\frac{a}{b}+\frac{c}{d}\right)\cdot\left(\frac{b}{c}+\frac{d}{a}\right)}=$$
$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)}=$$
$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{4+\left(\sqrt{\frac{a}{c}}-\sqrt{\frac{c}{a}}\right)^2+\left(\sqrt{\frac{b}{d}}-\sqrt{\frac{d}{b}}\right)^2}\ge$$
$$\ge 2\sqrt{4}=4.$$
A: I don't know whether the following argument will be considered as AM-GM (basically, this is the way one proves AM-GM for 4 terms)
Take $a=x^2, b=y^2, c=z^2, d=t^2$.
Then we have $\frac{a}{b}+\frac{d}{a}=\frac{x^2}{y^2}+\frac{t^2}{x^2}=\frac{x^2}{y^2}-2\frac{x}{y}\frac{t}{x}+\frac{t^2}{x^2}+2\frac{t}{y}=(\frac{x}{y}-\frac{t}{x})^2+2\frac{t}{y}\ge2\frac{t}{y}=2\frac{\sqrt{d}}{\sqrt{b}}.$  $\space\space\space\space\space\space\space\space (1)$
Now, if we replace $a, b, d$ in $(1)$ with respectively $c,d,b$, we'll get
$\frac{c}{d}+\frac{b}{c}\ge 2\frac{\sqrt{b}}{\sqrt{d}}$.
Next, use $(1)$ again, using numbers $\sqrt{b},\sqrt{d},\sqrt{d}$ and you'll get the desired result.
