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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.

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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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    $\begingroup$ This is black magic... $\endgroup$ – chubakueno Dec 23 '13 at 5:56
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    $\begingroup$ How is it humanly possible to derive this? How to derive the fact that $$\dfrac{x^2}{(x-1)^2}+\dfrac{y^2}{(y-1)^2}+\dfrac{z^2}{(z-1)^2}=\left(\frac{x}{x-1} +\dfrac{y}{y-1}+\dfrac{z}{z-1}-1\right)^2+1$$I can prove this by expanding, but there's no way on earth I would've found that equality myself. $\endgroup$ – user26486 Dec 23 '13 at 21:38
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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.

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  • $\begingroup$ I'm young and I haven't even started learning about calculus. So we'll have to find a different method for solving this. $\endgroup$ – user26486 Dec 23 '13 at 0:14
  • $\begingroup$ I wish you had said "without using calculus" in your post. $\endgroup$ – T.J. Gaffney Dec 23 '13 at 0:14
  • $\begingroup$ I'm sorry. At least you've probably got some more experience. $\endgroup$ – user26486 Dec 23 '13 at 0:26
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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 "="

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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*}

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  • $\begingroup$ I don't understand what you mean by $a,b,c$. I use $x,y,z$ in the problem. $\endgroup$ – user26486 Dec 23 '13 at 21:18

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