Ravi substitution: abc ≥ (a + b − c)(b + c − a)(c + a − b)

Ravi substitution. The triangle inequalities say exactly that $x,y,z>0$, and conversely $a=y+z$, $b=z+x$, $c=x+y$. So the sides of a triangle correspond to arbitrary positive numbers.

1. In terms of $x,y,z$ the inequality reads
$$(x+y)(y+z)(z+x)\ge(2z)(2x)(2y)=8xyz,$$
which is true by AM-GM on each factor: $x+y\ge2\sqrt{xy}$, etc. Equality iff $x=y=z$, i.e. the triangle is equilateral.

2. Since $b+c-a=2x$, etc.,
$$\sum\frac a{b+c-a}=\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}=\frac12\left[\left(\frac xy+\frac yx\right)+\left(\frac yz+\frac zy\right)+\left(\frac zx+\frac xz\right)\right]\ge\frac12\cdot6=3.$$

Lesson. A triangle condition is a hidden positivity condition: the Ravi substitution replaces it by three free positive variables, which makes many geometric inequalities routine.