Show that $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \geq x+y+z $ by considering homogeneity

Solution 1:

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.

Solution 2:

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 $$ x + y + z \leq \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} $$

Solution 3:

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.