<s>[INST] show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $ We want to show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq  \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $
I've tried to use AM-GM directly on the LHS but that obviously failed.
Simplifying the question gives $a^2b + b^2c + c^2a \geq ab + bc + ac$
The fact that this is $a(ab) + b(bc) + c(ca) \geq ab + bc + ac$ makes me think that there's a way to proceed from here, but I'm not quite sure how.
Note : a,b,c are real positive numbers
 [/INST] $$\begin{align}\frac{a}{b} + \frac{b}{c} + \frac{c}{a}
&\geq \frac{(abc)^{1/3}}{a} + \frac{(abc)^{1/3}}{b} + \frac{(abc)^{1/3}}{c}\\ 
&\geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
\end{align}$$
The second inequality is obvious while the first one comes from Karamata's Majorization Inequality with the convexity of $f(x)=e^x$ and
$$\left(\ln\dfrac{(abc)^{1/3}}{c}, \ln\dfrac{(abc)^{1/3}}{b}, \ln\dfrac{(abc)^{1/3}}{a}\right) \prec   \left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)
$$
Let (WLOG) $\;\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$
We have $\;\dfrac{a}{b}\ge\dfrac{b}{c}\ge\dfrac{c}{a}$ thus, $\;ac\ge b^2\;$, $a^2\ge bc\;$ and $\;ab\ge c^2\;$ hence, we get $a\ge b$ and $a\ge c$ thus,
$$\dfrac{a}{b}\ge \dfrac{(abc)^{1/3}}{c} \implies \ln\dfrac{a}{b}\ge \ln\dfrac{(abc)^{1/3}}{c}$$
$$\dfrac{c}{a}\le \dfrac{(abc)^{1/3}}{a} \implies \ln\dfrac{c}{a}\le \ln\dfrac{(abc)^{1/3}}{a}$$
$$0=\ln\dfrac{(abc)^{1/3}}{c}+ \ln\dfrac{(abc)^{1/3}}{b}+ \ln\dfrac{(abc)^{1/3}}{a}=   
\ln\dfrac{a}{b}+ \ln\dfrac{b}{c}+ \ln\dfrac{c}{a}
$$
so that our majorization rule holds, hence first inequality holds.
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