Another symmetric inequality How would one show that for positive $a,b,c,d$ and $a+b+c+d = 4$ that 
$$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \leq \frac{4}{abcd} $$
 A: Aryabhata's nice proof can be restated as :
$$ (ac+bd)((a+b+c+d)^4 - 64(abcc+bcdd+cdaa+dabb)) \\
=ac(16(ac+bd-ad-bc)^2+(a+b-c-d)^2((a+b+c+d)^2+4(a+b)(c+d))) \\
+bd(16(ac+bd-ab-cd)^2+(b+c-d-a)^2((a+b+c+d)^2+4(b+c)(d+a))) \\ \ge 0$$
Therefore, if $a+b+c+d = 4$, $abcc+bcdd+cdaa+dabb \le 4$.
A: Consider
$$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)abcd = a^2cd + b^2ad + c^2ab + d^2bc  = ac(ad + bc) + bd(ab + cd)$$
Since there is cyclic symmetry, we can assume that $ad + bc \le ab + cd$.
So
$$ac(ad + bc) + bd(ab + cd) \le (ac + bd)(ab + cd)$$
Now $xy \le \left(\frac{x+y}{2}\right)^2$
and so
$$(ac + bd)(ab + cd) \le \left(\frac{ac + bd + ab + cd}{2}\right)^2  = \left(\frac{(a+d)(b+c)}{2}\right)^2$$
Applying $xy \le \left(\frac{x+y}{2}\right)^2$ again we get
$$\left(\frac{(a+d)(b+c)}{2}\right)^2 \le \left(\frac{\left(\frac{a+b+c+d}{2}\right)^2}{2}\right)^2 = 4$$
Thus
$$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)abcd \le 4$$
and so
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \le \frac{4}{abcd}$$

What we have shown is that, for four positive numbers,
$$ \left(\frac{a+b+c+d}{4}\right)^4 \ge abcd\frac{\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d}+ \frac{d}{a}\right)}{4}$$
and since $\frac{a}{b} + \frac{b}{c} + \frac{c}{d}+ \frac{d}{a} \ge 4$, this inequality is stronger than $\text{AM} \ge \text{GM}$ for $4$ numbers. 
Somewhat surprisingly, we only used $\text{AM} \ge \text{GM}$ (twice)  to prove it! And for two numbers, a similar inequality is actually false!
A: Let's assume $a,b,c,d>0$. Rewriting your equation gives:
$$
\begin{eqnarray*}
a^2cd+b^2ad+c^2ab+d^2bc\leq 4
\end{eqnarray*}
$$
Equality is reached, if $a=b=c=d=1$. It's left to show, that this maximal:
Let $b=(2-a)$ with $0<a<2$ and $c=d=1$. Substituting this, gives
$$
\begin{eqnarray*}
(a-2)(a-1)^2&<&0
\end{eqnarray*}
$$
which is true for the given range of $a$.
