In solving cyclic inequalities, some solutions break the initial symmetry by imposing an order (e.g., median condition) to transform the expression into an asymmetric form that’s easier to prove. For example:
- For a triangle with sides $a, b, c$, prove:
$$ a^2 b \left( a - b \right) + b^2 c \left( b - c \right) + c^2 a \left( c - a \right) \geq 0. $$
One solution assumes $b \neq \operatorname{med}\left( a, b, c \right)$ and rewrites the expression as:
$$ \left( c + a - b \right) \left( c - a \right)^2 b - \left( b + c - a \right) \left( a - b \right) \left( b - c \right) c \geq 0. $$
- For real numbers $a, b, c \in [1, 2]$, prove:
$$ \left( a + b + c \right) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \leq 10. $$
A solution assumes $b = \operatorname{med}\left( a, b, c \right)$ and transforms it into:
$$ b \left( 2c - a \right) \left( 2a - c \right) + \left( c + a \right) \left( a - b \right) \left( b - c \right) \geq 0. $$
These solutions break cyclic symmetry to simplify proofs. Are there general techniques or theorems (e.g., in group theory or optimization) that formalize symmetry breaking for proving cyclic inequalities? How can such transformations be systematically applied to similar problems, like those in 1 or 2?
