# $x\le y \le z$; $\text{ }x \cdot y \cdot z = 1$; Prove that $(x+1) \cdot(z+1)>3$

Good morning,

Assuming that $x$,$y$,$z$ are positive real numbers and $x \le y \le z$ and $x \cdot y \cdot z = 1$.

How can I prove that $(x+1) \cdot (z+1) > 3$?

I know that $x$ has to be less than $1$ and $z$ greater than $1$ and I have already tried to rearrange that term but this didn't help me.

So do you have any ideas how to prove that?

• – Michael Rozenberg Sep 10 '17 at 9:58
• Thanks! I didn't see that question but the answers there helped me. – Rumpel Sep 10 '17 at 11:12
• You are welcome! – Michael Rozenberg Sep 10 '17 at 11:51

Let $y < z$
Set $y_1 = z_1 = \sqrt{yz}$
$(x+1)(z+1)> (x+1)(z_1+1)$
So, you must prove you task only for $z = y$ => $z=\frac{1}{\sqrt{x}}$, $x \le 1$
$(x+1)(z+1) = 1 + x + \sqrt{x} + \frac{1}{\sqrt{x}} \ge 3+x$
(by $t + \frac{1}{t} \ge 2$ )