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 :) 

Just by Hölder inequality;
$(ab^2+bc^2+ca^2)(\frac1a+\frac1b+\frac1c)\ge(a+b+c)^2\tag1$
If $(\frac1a+\frac1b+\frac1c)\le(a+b+c)$ we're done. Otherwise;
$(ab^2+bc^2+ca^2)(a+b+c)\ge(ab+bc+ca)^2=(\frac1a+\frac1b+\frac1c)^2\tag2$
since $(a+b+c)<(\frac1a+\frac1b+\frac1c)$ we're done again.
