<s>[INST] how to prove this inequality $(ab+bc+ac)^2 ≥ 3abc(a+b+c)$ 
Prove that if $a,b,c$ are non-negative real numbers, then
  $(ab + bc + ca)^2 \geq 3abc(a+b+c)$.

I tried to compute from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$.
 [/INST] 
$$a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\ge3a^2bc+3ab^2c+3abc^2$$
  $$a^2b^2+a^2c^2+b^2c^2-a^2bc-ab^2c-abc^2\ge0$$
  $$2a^2b^2+2a^2c^2+2b^2c^2-2a^2bc-2ab^2c-2abc^2\ge0$$
  $$(a^2b^2-2a^2bc+a^2c^2)+(a^2c^2-2abc^2+b^2c^2)+(b^2c^2-2ab^2c+a^2b^2)\ge0$$
  $$(ab-ac)^2+(ac-bc)^2+(ab-bc)^2\ge0$$
  the upper operation is revertable

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