Show that $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \geq x+y+z $ by considering homogeneity Well, I'm preparing for an undergrad competition that is held in April and because of that I've been trying to solve the inequalities I find on the internet. I found this problem:
$$\displaystyle  \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \geq x+y+z \text{ for all }  x,y,z \in \mathbb{R^+}$$
It's easy to show that this inequality holds by applying the AM-GM inequality three times like the following:
$$ \frac{xy}{z} + \frac{xz}{y} \geq 2 \sqrt{\frac{xy}{z} \cdot\frac{xz}{y}} = 2x$$
$$ \frac{xy}{z} + \frac{yz}{x} \geq 2 \sqrt{\frac{xy}{z} \cdot\frac{yz}{x}} = 2y$$
$$ \frac{zy}{x} + \frac{xz}{y} \geq 2 \sqrt{\frac{zy}{x} \cdot\frac{xz}{y}} = 2z$$
Then summing these inequalities and canceling a factor of $2$ at the end will give us the desired inequality.
But if you look at the problem from another point of view, if we substitute $x'=\lambda x$, $y'=\lambda y$ and $z' = \lambda z$ the inequality stays the same. So, I was wondering if there could be another way of solving this inequality and the inequalities that are homogenous like this one in a more systematic way that could be applied to a broader range of problems.
I decided to add a constraint $x^2+y^2+z^2=1$ to the problem because the inequality in the problem could be thought of as a function in the three variables $x,y,z$ in $\mathbb{R}^3$. So, it's sufficient to study this function on the unit sphere because homogeneity allows us to define it for other points of $\mathbb{R}^3$, but I have no idea how this could lead me anywhere.
 A: Here's one way to take advantage of the homogeneity.
By symmetry and homogeneity, one can assume without loss of generality that $z=\min(x,y,z)$ and that $z=1$. This reduces the problem to the following.
Proposition: If $x\ge1$ and $y\ge1$, then $\displaystyle \frac{x}{y}+\frac{y}{x}+xy \ge 1+x+y$.
Proof: Assume that $x\ge1$ and $y\ge1$. Then
$$x y=((x-1)+1)((y-1)+1)=$$ 
$$(x-1)+(y-1)+1+(x-1)(y-1)\ge x+y-1\textrm,$$
because $(x-1)(y-1)$ is non-negative.
The sum of a positive real number and its reciprocal is always at least 2, so
$$\frac{x}{y}+\frac{y}{x} \ge 2\textrm.$$
Combining these two inequalities gives the desired result.
A: Not really an answer to your question, but I did find another rather simple way to prove it
Given $x,y,z \in \mathbb{R}^+$, 
\begin{align}
0 &\leq (xy - xz)^2 \\
&= (xy)^2 + (xz)^2 - 2x^2yz \\
2x^2yz &\leq (xy)^2 + (xz)^2
\end{align}
and similarly for $2y^2xz$ and $2z^2xy$. Then adding, and dividing by 2 gives
$$
x^2yz + y^2xz + z^2xy \leq (xy)^2 + (xz)^2 + (yz)^2
$$
then dividing by $xyz$ on both sides gives
$$
x + y + z \leq \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x}
$$
A: I don't think there is any particular advantage to adding a constraint in this case, (or in general), as the original inequality is proved easily enough as you noticed.
Another way, it is direct from Cauchy-Schwarz inequality that:
$$LHS^2 = \left(\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \right)\left(\frac{xz}{y} + \frac{yz}{x} + \frac{xy}{z} \right) \ge (x+y+z)^2 = RHS^2$$
OTOH, adding the constraint $xyz = 1$ to the original problem, gives the equivalent inequality:
$$\frac1{x^2} + \frac1{y^2} + \frac1{z^2} \ge \frac1{xy}+\frac1{yz} + \frac1{zx}$$
which is obvious by Rearrangement Inequality, and seems equally easy compared to AM-GM or C-S approach.
I am sure there are many more ways to show the inequality in homogeneous form or with a suitable constraint.  There may be other problems where one approach is preferable, though in general it seems more of personal preference rather than either approach being easier / more systematic.  
A: Well, I found another proof of the famous Nesbitt's inequality by considering its homogeneity. So, I thought it might be worth sharing it on here for future reference and also to show how homogeneity can be employed to prove a famous inequality by what I've learned today from @Macavity and @SteveKass. I'm going to post it as an answer because it is too long for an edit to my question.
Theorem: For $a,b,c \in \mathbb{R}^+$ we have:
$$ \frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{c+a} \geq \frac{3}{2} $$
Proof:
The LHS stays the same if we substitute $a'=\lambda a$, $b' = \lambda b$ and $c' = \lambda c$. 
Suppose that $a'+b'+c'=1$. This implies $\lambda a + \lambda b + \lambda c = \lambda (a+b+c) = 1$, which gives us $\displaystyle \lambda = \frac{1}{a+b+c}$. 
In other words, that means for any $a,b,c \in \mathbb{R}^+$ we can restrict our attention only to the case $a+b+c=1$. Because for any other triple $(x,y,z)$ we can always scale it by $\displaystyle \lambda= \frac{1}{x+y+z}$ to get to that case.
So, from now on, let's assume, Without Loss Of Generality, that $a+b+c=1$.
Now let's prove the inequality. We start by manipulating the inequality from the LHS:
$$ \frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{c+a} \geq \frac{3}{2} \iff  \frac{c}{a+b} +1+ \frac{a}{b+c}+1 + \frac{b}{c+a}+1 \geq \frac{3}{2}+3$$
$$ \frac{a+b+c}{a+b} + \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} \geq \frac{9}{2} \iff \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \geq \frac{9}{2}  $$
The last statement is correct because we have assumed $a+b+c=1$.
It's trivial that $2=2(a+b+c)=(a+b)+(b+c)+(c+a)$. So, multiplying both sides of the inequality by $2$ we get:
$$((a+b)+(b+c)+(c+a))(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}) \geq 9$$
The last inequality can be checked easily by applying AM-GM inequality or Cauchy-Schwarz inequality. Q.E.D.
A: In general, you want to use a restriction to achieve homogeneity, not use homogeneity to achieve a restriction.
