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 :) 
 A: 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.
A: You need to show that 
$$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) \geq 0$$ 
It suffices to show this on any set of the form $0 \leq a,b,c,d \leq N$. By calculus (the "extreme value theorem") the function $(a + b + c + d)^3 - 16(abc + bcd + cda + dab)$ achieves its minimum at some $(a,b,c,d)$ in the set $0 \leq a,b,c,d \leq N$. I claim that this minimum has to occur when $a = b = c = d$.
Write $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (a + b + c + d)^3 - 16(a + b)cd - 16(c + d)ab$$ 
Note that by AM-GM, we have $16(c + d)ab \leq 16(c + d)({a + b \over 2})^2$. If we had $a \neq b$, we could replace $a$ and $b$ by ${a + b \over 2}$, leaving $c$ and $d$ constant, and we'd get a smaller value. So since $(a,b,c,d)$ is the minimum, this can't happen and we conclude that $a = b$. For similar reasons $c = d$, reversing the roles of the terms $16(a + b)cd$ and $16(c + d)ab$
Next, write $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (a + b + c + d)^3 - 16(b + d)ac - 16 (a + c)bd$$ Then arguing like above gives $b = d$ and $a = c$. Combining the above gives $a = b = c = d$, whereupon $(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (4a)^3 - 16(4a^3) = 0$. Since this is the minimum, the expression $(a + b + c + d)^3 - 16(abc + bcd + cda + dab)$ is nonnegative as needed. 
A: I posted this inequality on http://www.artofproblemsolving.com/ and I received a nice answer. This answer can be checked on the following link : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=494463 . 
A: This is a version of Maclaurin's inequality. We have
$$S_1=\frac{a+b+c+d}{4}$$
$$S_3=\frac{abc+bcd+cda+dab}{4}$$
and so
$$S_1 \geqslant (S_3)^{\frac13}$$
gives the required result.
A: As noted by Zarrax, it suffices to show that
$$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = 4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc$$
is nonnegative for all $a, b, c, d \ge 0$. Here I use the notation $\sum_{\mathrm{sym}}f(a,b ,c, d) := (1/4!) \sum_{\pi \in S_4} f(\pi(a), \pi(b), \pi(c), \pi(d))$ as $\pi$ ranges over all permutations of $\{a, b, c, d\}$.
But by AM-GM, $a^3 + b^3 + c^3 \ge 3 abc$ so that $\sum_{\mathrm{sym}}(a^3 + b^3 + c^3) \ge 3 \sum_{\mathrm{sym}}abc$ or just $\sum_{\mathrm{sym}} a^3 \ge \sum_{\mathrm{sym}} abc$. Similarly, AM-GM also gives $a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \ge 6 abc$, so that $\sum_{\mathrm{sym}} a^2 b \ge \sum_{\mathrm{sym}} abc$. Thus
$$4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc
= 4 \left(\sum_{\mathrm{sym}}a^3  - \sum_{\mathrm{sym}}abc\right) + 36 \left(\sum_{\mathrm{sym}} a^2 b - \sum_{\mathrm{sym}}abc\right) \ge 0.$$
