$$\frac{4}{abcd}\geq\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$$
Given: $a+b+c+d=4$ and $a$, $b$, $c$ abd $d$ are positives.
How to prove the above inequality using Arithmetic Geometric Mean Inequality?
I tried the following but, I am getting stuck after last step.
Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$.
Hence, by Rearrengement and AM-GM we obtain: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=\frac{1}{abcd}(a^2cd+b^2da+c^2ab+d^2bc)=$$ $$=\frac{1}{abcd}(a\cdot acd+b\cdot bda+c\cdot cab+d\cdot dbc)\leq$$ $$\leq\frac{1}{abcd}(x\cdot xyz+y\cdot xyt+z\cdot xzt+t\cdot yzt)=$$ $$=\frac{1}{abcd}(x^2yz+y^2xt+z^2xt+t^2yz)=\frac{1}{abcd}(xy+zt)(xz+yt)\leq$$ $$\leq\frac{1}{abcd}\left(\frac{xy+zt+xz+yt}{2}\right)^2=$$ $$=\frac{1}{4abcd}\left((x+t)(y+z)\right)^2\leq\frac{1}{4abcd}\left(\frac{x+y+z+t}{2}\right)^4=\frac{4}{abcd}.$$ Done!