The validity of normalization in homogeneous inequalities? I'm going through a book on inequalities right now, and the author describes normalization with the following example. 

Prove that $a^2 + b^2 + c^2 \ge ab + bc + ca$

Of course the fundamental proof of this inequality has already been discussed before, so it has only been taken up for the purpose of demonstrating normalization.
The author continues saying that we may use the additional condition $abc = 1$. The reasoning is explained below.
Suppose $abc = k^3$. Let $a = kx$ , $b = ky$ , $c = kz$. Then it follows that $xyz = 1$. Using the established substitutions we get the same inequality in different variables: $x^2 + y^2 + z^2 \ge xy + yz + zx$ but with the additional condition that $xyz = 1$. 
My problem is that we seem to lose generality here by bounding the variables in the inequality. Thus we prove the inequality bounded by a specific condition, but beyond that condition the result is uncertain. Could it be that since the new inequality has been proven with the condition, then by definition of our substitutions, the original inequality is also true (by the multiplicative law)? This seems to be the logical explanation, but I want to be certain. Any help is greatly appreciated.    
 A: We are given the following:
$$x^2+y^2+z^2 \geq xy+yz+xz \text{ for } xyz=1$$
Multiply both sides of this equation by $k^2$:
$$(kx)^2+(ky)^2+(kz)^2 \geq k^2xy+k^2yz+k^2xz=(kx)(ky)+(ky)(kz)+(kx)(kz)$$
Let $a=kx$, $b=ky$, and $c=kz$:
$$a^2+b^2+c^2 \geq ab+bc+ac$$
Thus, from the specific case that the author proved, we can derive the general case.
The book likely wanted you to figure this out on your own and as you see this kind of "specification" proof technique used in multiple proofs, you'll likely be more able to be able to see how this is true on your own.  However, it kind of takes practice and I'm still getting used to it, too.
A: Here is another version of the normalization principle which is slightly more general than the particular case discussed in OP:
Normalization Principle: Let $\mathbb{R}_+$ denote the set of all non-negative real numbers. Let the functions $f, g: \mathbb{R}_+^n \to \mathbb{R}$ be homogeneous of degree $\alpha$. Let $N: \mathbb{R}_+^n \to \mathbb{R}_+$ be homogeneous of degree $\beta$, $\beta \neq 0$, such that $f(\mathbf{y}) \geq g(\mathbf{y})$ for all $\mathbf{y} \in \mathbb{R}_+^n$ satisfying $N(\mathbf{y}) = k$ where $k$ is a fixed positive constant. Then $f(\mathbf{x}) \geq g(\mathbf{x})$ for all $\mathbf{x} \in \mathbb{R}_+^n$ satisfying $N(\mathbf{x}) > 0$.
Proof: Let $\mathbf{x} = (x_1, x_2, \dots, x_n)$ be any point in $\mathbb{R}_+^n$ such that $N(\mathbf{x}) > 0$.
Now let $t = \big(\frac{k}{k'} \big)^{\frac{1}{\beta}}$ where $k' = N(\mathbf{x})$, and consider the point $\mathbf{y} = t \mathbf{x}$. Since $N$ is homogeneous of begree $\beta$, it follows that $N(\mathbf{y}) = N(t \mathbf{x}) = t^\beta N(\mathbf{x}) = k$. Hence, it follows from the hypothesis that $f(\mathbf{y}) \geq g(\mathbf{y})$.
Now $f$ and $g$ being homogeneous of degree $\alpha$, it follows that $f(\mathbf{y}) \geq g(\mathbf{y}) \iff f(t \mathbf{x}) \geq g(t \mathbf{x}) \iff t^\alpha f(\mathbf{x}) \geq t^\alpha g(\mathbf{x}) \iff f(\mathbf{x}) \geq g(\mathbf{x})$, where the last implication follows because $t > 0$.
