Proof that $2(a^4+b^4+c^4)<(a^2+b^2+c^2)^2$ if and only if line segments of length $a$,$b$ and $c$ form a triangle

Question

Obviously I tried to somehow manipulate the triangle inequality $a<b+c$

$a^2<(b+c)^2$

Similarly,

$b^2<(a+c)^2$ and $c^2<(a+b)^2$

After adding these together:

$a^2+b^2+c^2<(a+b)^2+(a+c)^2+(b+c)^2$

But it seems I even work in the wrong direction, cuz I have to find the lower bound on the sum of squares, not the one on the top. So then I tried to rewrite the triangle inequality like that:

$c>a-b$

$b>a-c$

$a>b-c$

and work from here but I also didn't manage to get to the point, where I feel like I have this problem in my bag

Answer 1

Hint. For $a,b,c\in\mathbb R^+$, we have $$a^4 + b^4 + c^4 - 2\cdot \left(a^2b^2 + b^2c^2 + a^2c^2\right)=(a+b+c)(a+b-c)(a+c-b)(a-b-c)$$ Therefore, your inequality is equivalent to $$(a+b+c)(a+b-c)(c+a-b)(a-b-c)<0\\\iff (a+b+c)(a+b-c)(c+a-b)(b+c-a)>0$$ Can you finish? Why is it trivial that the sides of a triangle fulfill this inequality? Why is there no other option, but $a,b,c$ to be sides of a triangle?