Another symmetric inequality

How would one show that for positive $a,b,c,d$ and $a+b+c+d = 4$ that $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \leq \frac{4}{abcd} $$

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!

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 $a+b+c+d = 4$, $abcc+bcdd+cdaa+dabb \le 4$.