# Any ideas for this cyclic inequality? If a,b,c,d>0 and a+b+c+d=8, prove that a/(b^2+2b)+ b/(c^2+2c)+c/(d^2+2d)+\d/(a^2+2a)>=(16)/((a+c)(b+d)).

Any ideas for this cyclic inequality?
If $a,b,c,d>0$ and $a+b+c+d=8$, prove that
$\frac{a}{{b}^{2}+2b}+\frac{b}{{c}^{2}+2c}+\frac{c}{{d}^{2}+2d}+\frac{d}{{a}^{2}+2a}⩾\frac{16}{\left(a+c\right)\left(b+d\right)}.$
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By Holder and AM-GM we obtain:
$\sum _{cyc}\frac{a}{{b}^{2}+2b}=\frac{\sum _{cyc}ab\sum _{cyc}a\left(b+2\right)\sum _{cyc}\frac{a}{b\left(b+2\right)}}{\sum _{cyc}ab\sum _{cyc}a\left(b+2\right)}\ge \frac{\left(a+b+c+d{\right)}^{3}}{\left(a+c\right)\left(b+d\right)\sum _{cyc}\left(ab+2a\right)}=$
$=\frac{512}{\left(a+c\right)\left(b+d\right)\left(\left(a+c\right)\left(b+d\right)+16\right)}\ge \frac{512}{\left(a+c\right)\left(b+d\right)\left({\left(\frac{a+c+b+d}{2}\right)}^{2}+16\right)}=$
$=\frac{16}{\left(a+c\right)\left(b+d\right)}.$