$ \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$

Question

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$ ?

Answer 1

Using Holder's Inequality, we have:

$$\left(\sum_{cyc} \frac{a}{\sqrt[4]{16+b-d}}\right)^4 \left(\sum_{cyc} a(16+b-d) \right) \ge \left(\sum_{cyc} a \right)^5 = 16^5$$

$$\implies (LHS)^4 \times 16 \sum_{cyc} a \ge 16^5 \implies LHS \ge 8$$

Answer 2

By AM-GM and C-S $$\sum_{cyc}\frac{a}{\sqrt[4]{16+b-d}}=\sum_{cyc}\frac{32a}{4\cdot2^3\cdot\sqrt[4]{16+b-d}}\geq\sum_{cyc}\frac{32a}{3\cdot2^4+16+b-d}=$$ $$=\sum_{cyc}\frac{32a}{64+b-d}=\sum_{cyc}\frac{32a^2}{64a+ab-da}\geq\frac{32(a+b+c+d)^2}{\sum\limits_{cyc}(64a+ab-da)}=\frac{32\cdot16^2}{64\cdot16+0}=8.$$ Done!