show this inequality $(x+y)^3+(y+z)^3+(z+w)^3+(w+x)^3\ge 8(x^2y+y^2z+z^2w+w^2x)$

Question

let $x,y,z,w>0$,show that $$(x+y)^3+(y+z)^3+(z+w)^3+(w+x)^3\ge 8(x^2y+y^2z+z^2w+w^2x)$$

it seem use AM-GM inequality to solve it,But I can't it,Thanks

Answer 1

Let $x=\min\{x,y,z,w\}$, $y=x+a$, $z=x+b$ and $w=x+c$.

Thus, $$\sum_{cyc}\left((x+y)^3-8x^2y\right)=$$ $$=4(a^2+b^2+c^2-ab-bc)x+2(a^3+b^3+c^3)-5a^2b+3b^2a-5b^2c+3c^2b.$$ Id est, it's enough to prove that $$2(a^3+b^3+c^3)-5a^2b+3b^2a-5b^2c+3c^2b\geq0$$ or $$(2a^3-5a^2b+3ab^2+0.4b^3)+(1.6b^3-5b^2c+3bc^2+2c^3)\geq0.$$ Can you end it now?