$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c} \ge \frac{16}{a+b+c+d}$ for $a,b,c,d>0$

Question

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:

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Answer 1

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 $$ {1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq {16\over 3p} $$