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 :) 
 A: 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
$$x^2 \cdot x+ y^2 \cdot y + z^2 \cdot z \geq x^2 \cdot y+y^2 \cdot z+ z^2 \cdot x.$$
A: For fun, let us do this one also with Lagrange multipliers:
In[33]:= f = a b^2 + b c^2 + c a^2 - a - b - c

Out[33]= -a - b + a b^2 - c + a^2 c + b c^2

In[34]:= sol = Solve[
  {D[f, a] == k b c, D[f, b] == k a c, D[f, c] == k a b, a b c == 1, 
   a > 0, b > 0, c > 0},
  {a, b, c, k}
  ]

Out[34]= {{a -> 1, b -> 1, c -> 1, k -> 2}}

In[35]:= f /. sol[[1]]

Out[35]= 0

One has to check that my $f$ can only have a minimum, but that is easy.
A: To reiterate my comment above, Rearrangement Inequality needs some ordering in the variables, and hence cannot be applied here as the given inequality isn't symmetric wrt a,b,c.
To give a simple proof by AM-GM just note that $$a^2c + a^2c + ab^2 \ge 3 \sqrt[3]{a^5b^2c^2} = 3a$$ so adding the two other similar inequalities we get $ab^2 + bc^2 + ca^2 \ge a + b + c$.
A: 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.
