<s>[INST] Prove that $\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a$ 
Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d = 4$. Prove that $$\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a .$$

How can I approach this using only the AM - GM inequality? Are there any other methods that do not involve the AM-GM inequality?
 [/INST] This proof uses the rearrangement inequality in addition to AM-GM. After multiplying by $abcd$ our problem is equivalent to solving $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ Let $\{a,b,c,d\}=\{w,x,y,z\}$ with $w \ge x \ge y \ge z$. We have $$ a^2cd + ab^2d + abc^2 + bcd^2 = a(acd)+b(abd)+c(abc)+d(bcd)$$ and by the rearrangement equality $$ a(acd)+b(abd)+c(abc)+d(bcd) \le w(wxy)+x(wxz)+y(wyz)+z(xyz)$$ $$ = (wx+yz)(wy+xz)$$ By AM-GM, we get $$ (wx+yz)(wy+xz) \le \left(\frac{wx+yz+wy+xz}{2}\right)^2 = \frac{1}{4} \left((w+z)(x+y)\right)^2$$ Another AM-GM gives us $$ \frac{1}{4} ((w+z)(x+y))^2 \le \frac{1}{4} \left(\left(\frac{w+x+y+z}{2}\right)^2\right)^2 = \frac{1}{4} \left(\left(\frac{4}{2}\right)^2\right)^2 = 4$$ Thus we have $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ and furthermore, we have $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \le \frac{4}{abcd}$$ You might be able to tweak this to avoid the use of rearrangement, but I couldn't find a way.
</s>