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:
$$(a+1)(c+1)>3.$$
 A: 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 !
A: Since $b$ plays no role in the minimizing quantity, we might as well set $b=c$ to get the smallest $ac=\frac1b$. That means
$$
ac^2=1\tag{1}
$$
Using $(1)$, we want to find the minimum of
$$
\begin{align}
(a+1)(c+1)
&=\left(\frac1{c^2}+1\right)(c+1)\\
&=c+1+\frac1c+\frac1{c^2}\tag{2}
\end{align}
$$
$(2)$ is minimized when
$$
c^3-c-2=0\tag{3}
$$
The real root of $(3)$ is
$$
c=1.521379706805\tag{4}
$$
Plugging $(4)$ into $(2)$ says that the minimum is
$$
(a+1)(c+1)\ge3.610718613276\tag{5}
$$
A: 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!
