<s>[INST] When does this inequality hold? $\sum_{cyc}\frac{a^2+(a+b+d)c}{a^3+3bcd}\ge\frac{(a+b+c+d)^2}{a^3+b^3+c^3+d^3}$ 
For positive real numbers $a,b,c,d$, prove the inequality $$\frac{a^2+(a+b+d)c}{a^3+3bcd}+\frac{b^2+(a+b+c)d}{b^3+3acd}+\frac{c^2+(b+c+d)a}{c^3+3abd}+\frac{d^2+(a+c+d)b}{d^3+3abc}\ge\frac{(a+b+c+d)^2}{a^3+b^3+c^3+d^3}$$
When does equality hold?

I think that we have to use the $A.M-G.M$ inequality in some way here. I'm pretty sure we should start by simplifying the initial expression in some other way than multiplying.
I've thought about
$$a^3+bcd+bcd+bcd\ge4\sqrt[4]{a^3b^3c^3d^3}=4a^\frac{3}{4}b^\frac{3}{4}c^\frac{3}{4}d^\frac{3}{4}$$
$$a^2+ac+bc+cd\ge4\sqrt[4]{a^3bc^3d}=4a^\frac{3}{4}b^\frac{1}{4}c^\frac{3}{4}d^\frac{1}{4}$$
So
$$\frac{a^2+(a+b+d)c}{a^3+3bcd}=\frac{4a^\frac{3}{4}b^\frac{1}{4}c^\frac{3}{4}d^\frac{1}{4}}{4a^\frac{3}{4}b^\frac{3}{4}c^\frac{3}{4}d^\frac{3}{4}}=\frac{1}{\sqrt{bd}}$$
And we can repeat this for the other 3 expressions and get
$$\frac{2}{\sqrt{bd}}+\frac{2}{\sqrt{ac}}=\frac{a^2+(a+b+d)c}{a^3+3bcd}+\frac{b^2+(a+b+c)d}{b^3+3acd}+\frac{c^2+(b+c+d)a}{c^3+3abd}+\frac{d^2+(a+c+d)b}{d^3+3abc}$$
So now we have to prove
$$\frac{2}{\sqrt{bd}}+\frac{2}{\sqrt{ac}}\ge\frac{(a+b+c+d)^2}{a^3+b^3+c^3+d^3}$$
But I'm not sure how to do this, I think we have to use the $A.M-G.M$ inequality in some way again. Any help is appreciated.
 [/INST] Instead of solving LHS we will try to make from RHS
so RHS can be written as
$\frac{( a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd)}{( a^3 + b^3 + c^3 + d^3 )}$
which can be written as
$\frac{(a^2+c(a+b+d))}{(a^3+b^3+c^3+d^3)}$ + ..... similarly taking a square and 3 other terms
$a^3+b^3+c^3+d^3 \ge a^3+ (b^3\cdot c^3\cdot d^3)^\frac{1}{3}$
$a^3+b^3+c^3+d^3 \ge a^3 + 3abcd$
$\frac{(a^2+c(a+b+d))}{(a^3+b^3+c^3+d^3)} <= \frac{(a^2+c(a+b+d))}{(a^3+3abcd)}$ .. similarly for the other terms
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