Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $

Question

Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.

I know that i can use the axioms of the real numbers, but i can't finde an usefull equivalent transformation

Answer 1

multiplying your given inequality by $$xyz>0$$ we get $$x^2+y^2+z^2\geq xy+xz+zy$$ and this is equivalent to $$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$ which is true.

Answer 2

Multiply through by $xyz$ and you get $$x^2 + y^2 + z^2 \geq xy + yz + zx.$$ Now note that $$(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$ with equality only if $x = y = z$.

Answer 3

By C-S $$\sum_{cyc}\frac{x}{yz}=\sum_{cyc}\frac{x^2}{xyz}\geq\frac{(x+y+z)^2}{3xyz}\geq\frac{3(xy+xz+yz)}{3xyz}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

Answer 4

You can also use rearrangement inequality, which is true for all real numbers. The inequality is symmetric in terms of $x,y,z$. Let WLOG $x\ge y\ge z$. Notice that it's only given that $xyz>0$, $x,y,z\in\mathbb R$, not necessarily $x,y,z>0$. There are two cases:

$1)$ If $z>0$, then $y>0$, $x>0$.

$2)$ If $z<0$, then $y<0$, $x>0$.

In both cases $(x,y,z)$ and $\left(\frac{1}{yz},\frac{1}{xz},\frac{1}{xy}\right)$ are sorted in the same order.

$(x,y,z)$ and $\left(\frac{1}{xy},\frac{1}{yz},\frac{1}{zx}\right)$ is another way to sort them, but we don't care if they're sorted in the same or opposite or another order.

$$x\cdot\frac{1}{yz}+y\cdot\frac{1}{xz}+z\cdot \frac{1}{xy}\ge $$

$$\ge x\cdot\frac{1}{xy}+y\cdot\frac{1}{yz}+z\cdot \frac{1}{zx}=$$

$$=\frac{1}{y}+\frac{1}{z}+\frac{1}{x}$$

Edit: as other answers have noticed, multiplying both sides by $xyz>0$ gives $x^2+y^2+z^2\ge xy+yz+zx$ and there has been a proof shown. Another proof is using rearrangement inequality. $(x,y,z)$ and $(x,y,z)$ are sorted in the same order. $(x,y,z)$ and $(y,z,x)$ is another way to sort them.