$ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$ Let $a,b,c,d>0 ; a+b+c+d=16$ , then how to prove that 
$ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$ ?

Using Holder's Inequality, we have:
$$\left(\sum_{cyc} \frac{a}{\sqrt[4]{16+b-d}}\right)^4 \left(\sum_{cyc} a(16+b-d) \right) \ge \left(\sum_{cyc} a \right)^5 = 16^5$$
$$\implies (LHS)^4 \times 16 \sum_{cyc} a \ge 16^5 \implies LHS \ge 8$$
