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