for $abc = 1$ and $a \le b \le c$ prove that $(a+1)(c+1)>3$ This inequality has been given to me by my teacher to keep me occupied and after hours of fumbling around with it, and later trying to google it. I found nothing at all. For any three positive real numbers $a$, $b$ and $c$, where $abc = 1$ and $a\le b \le c$, prove that: $$(a+1)(c+1)>3.$$

First simplify the expression :

$ac + a + c + 1 \gt 3 \iff ac + a+c \gt 2 \iff 1/b + a+ c \gt 2 \iff 1 + ab+bc \gt 2b $

We should prove $1+ab+bc \gt 2b$ . Use the condition $a\le b\le c $ :

$a\le b\le c \iff 2a \le a+b \le a+c \to a+b\le a+c$

From these we have :

$a+b\le a+c \iff ab + b^2 \le ab + bc \iff ab + b^2 +1 \le ab + bc+ 1$

If we prove $ab + b^2 +1 \gt 2b $ then problem is solved :

$ab + b^2 +1 \gt 2b \iff ab + (b-1)^2 \gt 0$

which is obvious because $ab \gt 0 $ and $(b-1)^2 \gt 0$

Done !