<s>[INST] If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$. If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$$
Without using calculus.
There are a few ways I've tried solving this:
$1)$ We could try using the Cauchy-Schwarz inequality: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge \frac{(x+y+z)^2}{(x-1)^2+(y-1)^2+(z-1)^2}$$
But it's apparent that nothing's useful here.
$2)$ We could use AM-GM as well: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 3\sqrt[3]{\frac{x^2y^2z^2}{(x-1)^2(y-1)^2(z-1)^2}}=\frac{3}{\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}}$$
So we only have to prove that: $$\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}\le 3$$
We could raise both sides to the power of 3: $$(x-1)^2(y-1)^2(z-1)^2\le 27$$
But this inequality doesn't hold. 
$3)$ We could try cleaning the denominators by multiplying both sides by $(x-1)^2(y-1)^2(z-1)^2$. After a bunch of expanding and simplifying we get that: $$x^2y^2+y^2z^2+x^2z^2-6(xy+yz+xz)+2(x+y+z)+9\ge 0$$
I can't tell so easily whether the inequality is true or not. You could help me out on this one. Just don't forget that $x,y,z\in\mathbb R\setminus\{1\}$ and so we can't just simply use AM-GM, unless we're using it for squares that have to be non-negative.
 [/INST] Other nice solution:
since
$$\dfrac{x^2}{(x-1)^2}+\dfrac{y^2}{(y-1)^2}+\dfrac{z^2}{(z-1)^2}=\left(\dfrac{x}{x-1}
+\dfrac{y}{y-1}+\dfrac{z}{z-1}-1\right)^2+1\ge 1$$
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