<s>[INST] For any triangle with sides $a$, $b$, $c$, show that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$ 
For any triangle with sides $a$, $b$, $c$, prove the inequality $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0 .$$

This is IMO 1983 problem 6.
I tried substituting $a=x+y$, $b=y+z$, $c=z+x$ but well it doesn't help in any sense except wasting 3 pages that lead to nothing (please don't mind the joke). Using $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ also didn't lead to anything for me. Could you give me a hint for finding the proper substitution?
 [/INST] $$ \it{a^2 b(a-b)+b^2c(b-c)+c^2 a(c-a)}$$
$$ \it{={\frac {bc\, \left( a+b-c \right) ^{2} \left( a-b \right) ^{2}+a \left( c+a-b \right)  \left\{ b \left( a+b-c \right)  \left( a-c \right) ^{2} +a \left( b+c-a \right)  \left( b-c \right) ^{2} \right\} }{a \left( c +a-b \right) +b \left( a+b-c \right) }}}$$
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