$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c} \ge \frac{16}{a+b+c+d}$ for $a,b,c,d>0$ Prove that
$$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c}
 \ge \frac{16}{a+b+c+d}$$
if $a,b,c,d>0$.
My attempt:
I've put this in the answers.
See also comments under this question.

My attempt:
Let p=a+b+c+d then
$$
\frac{1}{b+c+d}+ \frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\equiv\\{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d}
$$
By the AM-GM inequality,
$$
{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq4\sqrt[4]{1\over(p-a)(p-b)(p-c)(p-c)}.
$$
Also by AM-GM inequality,
$$
\sqrt[4]{(p-a)(p-b)(p-c)(p-c)}\leq\frac14 \left((p-a)+(p-b)+(p-c)+(p-d)\right)\\
=\frac14(4p-p)=\frac34p\\
$$
so
$$
{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq {16\over 3p}
$$
