<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] The Ravi substitution that you did works just fine.
With $a = y+z,b = x+z$ and $c = x+y$, we will need to prove:
$$\sum\limits_{cyc} (y+z)^2(z+x)(y-x) \ge 0$$
Simplifies to, $xy^3 + yz^3 + zx^3 \ge x^2yz + xy^2z + xyz^2$
Use Cauchy-Schwarz Inequality:
$\displaystyle \begin{align}x^2yz + xy^2z + xyz^2 &= \sum\limits_{cyc} (x^{3/2}z^{1/2})(x^{1/2}yz^{1/2}) \\&\le (x^3z + y^3x + z^3y)^{1/2}(xy^2z + xyz^2 + x^2yz)^{1/2}\end{align}$
Thus, $x^2yz + xy^2z + xyz^2 \le xy^3 + yz^3 + zx^3$
The link provided by @math110 gives a simpler explanation:
Rewrite the inequality as:
$\displaystyle \sum\limits_{cyc} \frac{y^2}{z} \ge x+y+z$ which is Engel's form $\displaystyle \left(\sum\limits_{cyc} \frac{y^2}{z}\right)(x+y+z) \ge (x+y+z)^2$
or we can use Rearrangement as @math110 suggests.
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