<s>[INST] Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ I want to prove the following inequality : 

$$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab,  $$
  $a,b,c,d \in \mathbb{R}_{+} .$ 

In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that. 
Thanks :) 
 [/INST] Also notice that:
$$(a+b+c+d)^3 - 16(abc+abd+acd+bcd) = (a+b+c+d)(a+b-c-d)^2 + 4(c-d)^2(a+b) + 4(a-b)^2(c+d) \ge 0$$
Or
$$(a+b)[(a+b-c-d)^2 + 4(c-d)^2] + (c+d)[(a+b-c-d)^2 + 4(a-b)^2]\ge0$$
This way is suggested by a friend of mine.
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