<s>[INST] 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:
$$(a+1)(c+1)>3.$$
 [/INST] Our conditions give $c\geq1$ and $c\geq b$.
Thus, by AM-GM we obtain:
$$(a+1)(c+1)=ac+a+3\cdot\frac{c}{3}+1\geq5\sqrt[5]{ac\cdot a\cdot\left(\frac{c}{3}\right)^3}+1=$$
$$=\frac{5}{\sqrt[5]{27}}\sqrt[5]{a^2c^4}+1\geq\frac{5}{\sqrt[5]{27}}\sqrt[5]{a^2b^2c^2}+1=\frac{5}{\sqrt[5]{27}}+1>3.$$
Done!
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