Inequality. $ab^2+bc^2+ca^2 \geq a+b+c.$ Using rearrangement inequalities prove the following inequality: 
Let $a,b,c$ be positive real numbers satisfying $abc=1$. Prove that 
$$ab^2+bc^2+ca^2 \geq a+b+c.$$
Thanks :) 

I think I have the solution using arrangements inequalities.(source: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=497213) 
We make the substitution $\displaystyle a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$. We have now: 
$$\frac{z^2}{xy}+\frac{x^2}{yz}+\frac{y^2}{xz} \geq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}.$$
So: 
$$z^3+x^3+y^3\geq y^2z+z^2x+x^2y.$$ And this inequality can be solved using rearrangements inequality.
Let $x \geq y \geq z$. Using rearrangement inequality for $(x^2,y^2,z^2)$ and $(x,y,z)$ we conclude that
$$x^2 \cdot x+ y^2 \cdot y + z^2 \cdot z \geq x^2 \cdot y+y^2 \cdot z+ z^2 \cdot x.$$
