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


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 $$2(a^3+b^3+c^3)-5a^2b+3b^2a-5b^2c+3c^2b\geq0.$$ Can you end it now?

  • $\begingroup$ Dear Michael, v. nice solution, once more, but how on Earth do you solve problems so fast? $\endgroup$ – math_here Jun 12 at 15:35
  • $\begingroup$ I just like to prove inequalities. $\endgroup$ – Michael Rozenberg Jun 12 at 15:37
  • $\begingroup$ Based on how good you are, you not only like it, you love it :) $\endgroup$ – math_here Jun 12 at 15:39

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