Prove that if $(a^2+b^2+c^2+d^2)^2 > 3(a^4+b^4+c^4+d^4)$, then, using any three of them we can construct a triangle. 
Prove that if $a,b,c,$ and $d$ are positive numbers and satisfy $(a^2+b^2+c^2+d^2)^2 > 3(a^4+b^4+c^4+d^4)$, then, using any three of them we can construct a triangle.

I find it hard to go from the given inequality to the triangle inequality for $3$ of $a,b,c,d$. Expanding it may help but that would get ugly. I did it and got $d^2 (2 a^2+2 b^2+2 c^2-2 d^2)+b^2 (2 a^2-2 b^2+2 c^2)+a^2 (2 c^2-2 a^2)-2 c^4$. Also doing a ravi substitution here seems almost impossible to do with $4$ variables.

We'll prove that $a$, $b$ and $c$ are sides-lengths of a triangle.
Indeed, by C-S 
$$3(a^4+b^4+c^4+d^4)=(2+1)(a^4+b^4+c^4+d^4)\geq\left(\sqrt{2(a^4+b^4+c^4)}+d^2\right)^2$$
which gives $a^2+b^2+c^2>\sqrt{2(a^4+b^4+c^4)}$ or
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)>0$$
and we are done because $a+b-c<0$ and $a+c-b<0$  impossible. 
