Why is the triangle inequality equivalent to a^4+b^4+c^4 <= 2(a^2 b^2+b^2 c^2+c^2 a^2)

Why is the triangle inequality equivalent to $a^4+b^4+c^4\le <!--\; \le \; -->2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$
Consider the existential problem of a triangle with side lengths $a,b,c\ge <!--\; \ge \; -->0$. Such a triangle exists if and only if the three triangle inequalities
$\begin{array}{}\text{(0)}& a+b\ge <!--\; \ge \; -->c,b+c\ge <!--\; \ge \; -->a\text{and}c+a\ge <!--\; \ge \; -->b\end{array}$
are all satisfied.
Alternatively, if ${\mathrm{\ell <!--\; \ell \; -->}}_1\le <!--\; \le \; -->{\mathrm{\ell <!--\; \ell \; -->}}_2\le <!--\; \le \; -->{\mathrm{\ell <!--\; \ell \; -->}}_3$ are the values of a,b and c ordered in ascending order, then the triangle exists iff ${\mathrm{\ell <!--\; \ell \; -->}}_1+{\mathrm{\ell <!--\; \ell \; -->}}_2\ge <!--\; \ge \; -->{\mathrm{\ell <!--\; \ell \; -->}}_3$.
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let $0,x,y\in <!--\; \in \; -->{\mathbb{R}}^2$ be the three vertices of the triangle, with $\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->=a,\Vert <!--\; \Vert \; -->y\Vert <!--\; \Vert \; -->=b$ and $\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y\Vert <!--\; \Vert \; -->=c$. Then $c^2=\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y{\Vert <!--\; \Vert \; -->}^2=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}^2=a^2+b^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->$
Therefore $x^{T}y=⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->=\frac{1}{2}(a^2+b^2-<!--\; -\; -->c^2)$ and
$\begin{array}{}\text{(1)}& \left(\begin{array}{c}{x}^T\\ y^T\end{array}\right)\left(\begin{array}{cc}x& y\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}2a^2& a^2+b^2-<!--\; -\; -->c^2\\ a^2+b^2-<!--\; -\; -->c^2& 2b^2\end{array}\right).\end{array}$
The RHS of (1) must be positive semidefinite because the LHS is a Gram matrix. Conversely, if the RHS is indeed PSD, it can be expressed as a Gram matrix. Hence we obtain x and y and the triangle exists.
As $2a^2$ and $2b^2$ are already nonnegative, the RHS of (1) is positive semidefinite if and only if $(2a^2)(2b^2)-<!--\; -\; -->(a^2+b^2-<!--\; -\; -->c^2{)}^2\ge <!--\; \ge \; -->0$, by Sylvester's criterion. That is, the triangle exists if and only if
$\begin{array}{}\text{(2)}& -<!--\; -\; -->(a^4+b^4+c^4)+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)\ge <!--\; \ge \; -->0.\end{array}$
This polynomial inequality can be derived by more elementary means. See circle-circle intersection on Wolfram MathWorld. The geometric explanation for the necessity of (2) is given by Heron's formula, which states that the square root of the LHS is four times the area of the triangle.
Since both (0) and (2) are necessary and sufficient conditions for the existence of the required triangle, the two sets of conditions must be equivalent to each other. Here are my questions. Is there any simple way to see why (0) and (2) are equivalent? Can we derive one from the other by some basic algebraic/arithmetic manipulations?