Any ideas for this cyclic inequality?

Question

If $a,b,c,d>0$ and $a+b+c+d=8$, prove that $$\frac{a}{b^2+2b}+ \frac{b}{c^2+2c}+\frac{c}{d^2+2d}+\frac{d}{a^2+2a}\geqslant\frac{16}{(a+c)(b+d)}.$$

I tried to use Titu Andreescu's inequality, but i didn't got anywhere. Can anyone give me another idea?

Answer 1

By Holder and AM-GM we obtain: $$\sum_{cyc}\frac{a}{b^2+2b}=\frac{\sum\limits_{cyc}ab\sum\limits_{cyc}a(b+2)\sum\limits_{cyc}\frac{a}{b(b+2)}}{\sum\limits_{cyc}ab\sum\limits_{cyc}a(b+2)}\geq\frac{(a+b+c+d)^3}{(a+c)(b+d)\sum\limits_{cyc}(ab+2a)}=$$ $$=\frac{512}{(a+c)(b+d)((a+c)(b+d)+16)}\geq\frac{512}{(a+c)(b+d)\left(\left(\frac{a+c+b+d}{2}\right)^2+16\right)}=$$ $$=\frac{16}{(a+c)(b+d)}.$$ I used the following Holder for three sequences.

Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $c_1,$ $c_2,$... $c_n$, $\alpha$, $\beta$ and $\gamma$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}(c_1+c_2+,,,+c_n)^{\gamma}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}c_1^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+\left(a_2^{\alpha}b_2^{\beta}c_2^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+...+\left(a_n^{\alpha}b_n^{\beta}c_n^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}\right)^{\alpha+\beta+\gamma}$$