Prove inequality: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ Prove: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ with $a, b, c \ge0$

I can do this by: $EAT^2$ (expand all of the thing)


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*$(x+y+z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y+4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2}+4\,x{y}^{3}+4\,x{z}^{3}+4\,{y}^{3}z+6\,{y}^{2}{z}^{2}+4
\,y{z}^{3}+12\,x{y}^{2}z+12\,xy{z}^{2}+12\,{x}^{2}yz$

*$(x+y-z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y-4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2}+4\,x{y}^{3}-4\,x{z}^{3}-4\,{y}^{3}z+6\,{y}^{2}{z}^{2}-4
\,y{z}^{3}-12\,x{y}^{2}z+12\,xy{z}^{2}-12\,{x}^{2}yz$
...
$$28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4\\
 \iff a^4 + b^4 + c^4 \ge a^2b^2+c^2a^2+b^2c^2 \text{(clearly hold by AM-GM)}$$ 
but any other ways that smarter ?

A nice way of tackling the calculations might be as follows:$$~$$
Let $x=b+c-a,y=c+a-b,z=a+b-c.$ Then the original inequality is just equivalent with
$$\frac74\Bigl((x+y)^4+(y+z)^4+(z+x)^4\Bigr)\geq x^4+y^4+z^4+(x+y+z)^4.$$
Now we can use the identity
$$\sum_{cyc}(x+y)^4=x^4+y^4+z^4+(x+y+z)^4-12xyz(x+y+z),$$
So that it suffices to check that
$$\frac37\Bigl(x^4+y^4+z^4+(x+y+z)^4\Bigr)\geq 12xyz(x+y+z),$$
Which obviously follows from the AM-GM inequality: $(x+y+z)^4\geq \Bigl(3(xy+yz+zx)\Bigr)^2\geq 27xyz(x+y+z)$ and $x^4+y^4+z^4\geq xyz(x+y+z).$
Equality holds in the original inequality iff $x=y=z\iff a=b=c.$
$\Box$
