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.
 A: I think I got it:
Substitute $z=\frac{1}{xy}$.  Take the left hand side of the inequality and minimize it using calculus.  We will show that even when minimum, it will be greater than $1$.
When you minimize and reduce, you should get $-x(1-xy)^3+y(x-1)^3=0$ and $-y(1-xy)^3+x(y-1)^3=0$.  Solving we get, $\frac{x^2}{(x-1)^2}=\frac{(x-1)}{(y-1)}\frac{y^2}{(y-1)^2}$.  If $x-1$ and $y-1$ are the same sign, then assume WLOG $x-1>y-1$.  When you substitute back in you get $\frac{x-1}{y-1}\left(1+\frac{y^2}{(y-1)^2}\right)+\left(\frac{1}{1-xy}\right)^2$, which is greater than 1.
What if $x-1$ and $y-1$ are not the same sign, then $z-1$ must be the same sign as one of those two, so start over that way.
A: Other nice solution:
since
$$\dfrac{x^2}{(x-1)^2}+\dfrac{y^2}{(y-1)^2}+\dfrac{z^2}{(z-1)^2}=\left(\dfrac{x}{x-1}
+\dfrac{y}{y-1}+\dfrac{z}{z-1}-1\right)^2+1\ge 1$$
A: Proof:
Use Cauchy-Schwarz inequality,we have
$$\left[\sum_{cyc}\left(\dfrac{a}{a-b}\right)^2\right]\left[\sum_{cyc}(a-b)^2(a-c)^2\right]\ge\left(\sum_{cyc}a^2-\sum_{cyc}ab\right)^2$$
and since
\begin{align*}
\sum_{cyc}(a-b)^2(a-c)^2&=\sum_{cyc}(a-b)^2(a-c)^2+2\sum_{cyc}(a-b)(a-c)(b-c)(b-a)\\
&=\left[\sum_{cyc}(a-b)(a-c)\right]^2=\left(\sum_{cyc}a^2-\sum_{cyc}ab\right)^2
\end{align*}
A: for your last inequality, note:
$(xy+yz+xz)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)=x^2y^2+y^2z^2+z^2x^2+2(x+y+z)$
let $xy+yx+xz=t \ge 3 \sqrt[3]{(xyz)^2}=3$
LHS$=t^2-6t+9=(t-3)^2\ge 0 $  when $t=3$ hold "="
