Prove $(a^2+b^2+c^2+d^2)^2≥(a+b)(b+c)(c+d)(d+a)$ I've been unsuccessfully trying to solve this contest-style problem for a while. 
Tried different substitutions and the such, but nothing helped. I presume the solution is related to Cauchy-Schwarz? Anyway, any hints would be appreciated!

Given $a, b, c, d \in \mathbb{R^+}$, prove that
  $$(a^2 + b^2 + c^2 + d^2)^2 \ge (a+b)(b+c)(c+d)(d+a)$$

Note- I'm new on the site, sorry if I'm breaking some rule or convention.

By the AM-GM inequality:
$$ (a+b)(c+d) \leq \frac{1}{4}(a+b+c+d)^2, $$
$$ (b+c)(d+a) \leq \frac{1}{4}(a+b+c+d)^2, $$
and by the AM-QM inequality:
$$ \frac{1}{4}(a+b+c+d)^2 \leq (a^2+b^2+c^2+d^2).$$
The claim easily follows.
