<s>[INST] Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq1+4\sqrt{\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq1+4\sqrt{\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$
We can prove this inequality by using $b=a+u$, $c=a+v$ and $d=a+w$ after squaring of the both sides.
I am looking for a nice proof. Thank you!
 [/INST] Let:  $ a+b+c+d=1 $
Then the inequality can be rewritten:
$$a\cdot \frac{c+d}{a+b}+b \cdot \frac{d+a}{b+c}+c \cdot \frac{a+b}{c+d}+d \cdot \frac{b+c}{d+a} \ge 4 \cdot \sqrt {\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$
$$\frac{a (c+d)^2+c (a+b)^2}{(a+b)(c+d)}+\frac{b (d+a)^2+d (b+c)^2}{(b+c)(d+a)} \ge 2 \cdot \sqrt {\frac{(a (c+d)^2+c (a+b)^2)(b (d+a)^2+d (b+c)^2)}{(a+b)(b+c)(c+d)(d+a)}} \ge$$ 
$$\ge  4\cdot \sqrt {\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$
$$\Leftrightarrow (a (c+d)^2+c (a+b)^2)(b (d+a)^2+d (b+c)^2) \ge 4abcd$$
$$4 abcd \sum\limits_{cyc} a^2+  \left ( 5 abcd\sum\limits_{cyc}ab+\sum\limits_{cyc}(a^4bc+ab^2d^3+ac^2d^3) \right) + $$
$$+\left ( 4abcd \cdot (ac+bd)+2 (a^3b^2c+ac^3d^2)+2 (b^3c^2d+bd^3a^2) \right) \ge 4abcd\cdot  (a+b+c+d)^2= 4abcd$$
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