Why is the triangle inequality equivalent to $a^4+b^4+c^4\leq 2(a^2b^2+b^2c^2+c^2a^2)$? Consider the existential problem of a triangle with side lengths $a,b,c\geq0$. Such a triangle exists if and only if the three triangle inequalities
$$a+b\geq c,\quad b+c\geq a\quad\text{and}\quad c+a\geq b\tag{0}$$
are all satisfied.
Alternatively, if $\ell_1\leq\ell_2\leq\ell_3$ are the values of $a,b$ and $c$ ordered in ascending order, then the triangle exists iff $\ell_1+\ell_2\geq\ell_3$.
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let $0,x,y\in\mathbb R^2$ be the three vertices of the triangle, with $\|x\|=a,\,\|y\|=b$ and $\|x-y\|=c$. Then $c^2=\|x-y\|^2=\|x\|^2-2\langle x,y\rangle+\|y\|^2=a^2+b^2-2\langle x,y\rangle$. Therefore $x^Ty=\langle x,y\rangle=\frac{1}{2}(a^2+b^2-c^2)$ and
$$\pmatrix{x^T\\ y^T}\pmatrix{x&y}=\frac{1}{2}\pmatrix{2a^2&a^2+b^2-c^2\\ a^2+b^2-c^2&2b^2}.\tag{1}$$
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 $(2a^2)(2b^2)-(a^2+b^2-c^2)^2\geq0$, by Sylvester's criterion. That is, the triangle exists if and only if
$$-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\geq0.\tag{2}$$
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?
 A: The quartic polynomial in (2) may be shown by algebra to equal the product
$(a+b+c)(-a+b+c)(a-b+c)(a+b-c),$
so for nonnegative $a,b,c$ the triangle inequality implies that the quartic expression is nonnegative.
But the quartic as defined above is also nonnegative if two or all four of the factors above are negative. So we have to exclude these alternatives to prove equivalence.
Assume that $a+b-c$ and $a-b+c$ are negative. Simply adding these up gives $2a<0$, contradicting the hypothesis that $a$ is nonnegative. Similar contradictions occur if we try other pairs of factors being negative:
$(-a+b+c<0, a-b+c<0) \to (c<0)$
$(-a+b+c<0, a+b-c<0) \to (b<0)$
Thus by indirect proof the equivalence of (0) and (2) is established.
A: For non-negative $a,b,c$:
$$a+b\geq c; a+c\geq b;b+c\geq a \Leftrightarrow |b-a|\leq c \leq b+a\Leftrightarrow\\
(b-a)^2\leq c^2 \leq (b+a)^2 \Leftrightarrow (c^2-(b+a)^2)(c^2-(b-a)^2)\leq 0 \Leftrightarrow\\
c^4-((b+a)^2+(b-a)^2)c^2+(b^2-a^2)^2\leq 0\Leftrightarrow
c^4+b^4+a^4\leq 2(a^2b^2+a^2c^2+b^2c^2)$$
A: The equations in (0) can be equivalently expressed as
$$a + b - c \ge 0$$
$$a - b + c \ge 0$$
$$-a + b + c \ge 0$$
Since the product of non-negative factors must also be non-negative, we must have:
$$(a + b - c)(a - b + c)(-a + b + c) \ge 0$$
Of course, this product would also be positive if exactly two of the original three equations were false.  To prove equivalence, we must show that this is impossible.  Suppose that the first two were violated:
$$a + b - c < 0$$
$$a - b + c < 0$$
Adding up the two inequalities gives $2a < 0$, which contradicts the assumption that all of a triangle's sides are positive.  By symmetry, this contradiction occurs regardless of which two inequalities are chosen to be false.  Anyhow, multiplying everything out gives:
$$(a + b - c)(a - b + c)(-a + b + c) \ge 0$$
$$(a^2 - ab + ac + ab - b^2 + bc - ac + bc - c^2)(-a + b + c) \ge 0$$
$$(a^2 - b^2 + 2bc - c^2)(-a + b + c) \ge 0$$
$$-a^3 + a^2b + a^2c + ab^2 - b^3 - b^2c - 2abc + 2b^2c + 2bc^2 + ac^2 - bc^2 - c^3 \ge 0$$
$$-a^3 - b^3 - c^3 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - 2abc \ge 0$$
So now, we have a form of the triangle inequality that's a cubic polynomial.  It's a bit verbose, though, so let's see what happens if we include one more factor, $a + b + c \ge 0$.
$$(-a^3 - b^3 - c^3 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - 2abc)(a + b + c) \ge 0$$
$$-a^4 - a^3b - a^3c - ab^3 - b^4 - b^3c - ac^3 - bc^3 - c^4 + a^3b + a^2b^2 + a^2bc + a^2b^2 + ab^3 + ab^2c + a^3c + a^2bc + a^2c^2 + a^2c^2 + abc^2 + ac^3 + ab^2c + b^3c + b^2c^2 + abc^2 + b^2c^2 + bc^3 - 2a^2bc - 2ab^2c - 2abc^2 \ge 0$$
$$-a^4 - b^4 - c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2 \ge 0$$
Wow, that's a lot of intermediate terms that conveniently cancel out!  Now we can just rearrange this a little.
$$2(a^2b^2 + a^2c^2 + b^2c^2) \ge a^4 + b^4 + c^4$$
$$a^4 + b^4 + c^4 \le 2(a^2b^2 + a^2c^2 + b^2c^2)$$
Q.E.D.
