Let a,b,c,d>0 and $a+b+c+d=4$, prove $$\frac{4}{abcd}⩾\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$$ I found posts about this inequality, but I can't figure out where I'm wrong in my proof because I haven't read mine anywhere yet: $\frac{4}{abcd}⩾\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$
$4⩾(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a})(abcd)$
For AM-GM: $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}⩾4$
So $4⩾(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a})(abcd)⩾4(abcd)$.
This implies 1⩾abcd.
Claim: If $a+b+c+d=4$, then $1⩾abcd$
Proof: It's true by AM-GM
For positive variables we need to prove that: $$a^2cd+b^2ad+c^2ab+d^2bc\leq4.$$ Now, let $\{a,b,c,d\}=\{x,y,z,t\},$ where $x\geq y\geq z\geq t$.
Thus, by Rearrangement and AM-GM twice we obtain: $$a^2cd+b^2ad+c^2ab+d^2bc=a\cdot acd+b\cdot bad+c\cdot cab+d\cdot dbc\leq$$ $$\leq x\cdot xyz+y\cdot xyt+z\cdot xzt+t\cdot yzt=xy(xz+yt)+zt(xz+yt)=$$ $$=(xz+yt)(xy+zt)\leq\left(\frac{xz+yt+xy+zt}{2}\right)^2=$$ $$=\left(\frac{(x+t)(y+z)}{2}\right)^2\leq\left(\frac{\left(\frac{x+t+y+z}{2}\right)^2}{2}\right)^2=4.$$