<s>[INST] Another symmetric inequality How would one show that for positive $a,b,c,d$ and $a+b+c+d = 4$ that 
$$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \leq \frac{4}{abcd} $$
 [/INST] 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!
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