Note that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$. $$\sum_{cyc}\frac{a}{b}\sum_{cyc}\frac{ab}{a+b}\geq \frac{1}{2}\sum_{cyc} (a+b)\sum_{cyc}\frac{ab}{a+b}\overset{\text{C-S}}{\geq}\frac{(\sum_{cyc}\sqrt{ab})^2}{2}\overset{\text{AM-GM}}{\geq} \frac{9}{2}$$Equality holds when $a=b=c=1$.

By AM-GM $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} =\frac{1}{3}\left(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\right)+ \frac{1}{3}\left(\frac{b}{c}+\frac{b}{c}+\frac{c}{a} \right)+\frac{1}{3}\left(\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\right) \geq a+b+c$$$$ \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) \cdot \left(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}\right) \geq \frac{1}{2}((a+b)+(b+c)+(c+a))\cdot \left(\frac{1}{c(a+b)} + \frac{1}{a(b+c)} + \frac{1}{b(c+a)}\right)$$$$ \geq\frac{9}{2}\sqrt[3]{(a+b)(b+c)(c+a)\cdot\frac{1}{c(a+b)a(b+c)b(c+a)}}\geq\frac{9}{2}$$Equality holds when $a=b=c=1$.

Alternatively, note that the second multiplier is $\sum \frac{1}{ac+bc} \geq \frac{9}{2(ab+bc+ca)}$ by Titu's Lemma and that $a^2c + b^2a + c^2b \geq ab+bc+ca$ by summing cyclically the AM-GMs $a^2c + a^2c + c^2b \geq 3ac$.

Assassino9931 wrote: Let $a$, $b$ and $c$ be positive real numbers with $abc = 1$. Determine the minimum possible value of $$ \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right)\left(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}\right) $$as well as all triples $(a,b,c)$ which attain the minimum. The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers with $abc=1$. Prove that: $$ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{a^2b^2}{a^2+b^2}+\frac{b^2c^2}{b^2+c^2}+\frac{c^2a^2}{c^2+a^2}\right)\geq\frac{9}{2}.$$

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$$\sum \frac{a}{b} = \sum \frac{a^2c}{abc}= \sum a^2c \ge ab+bc+ac$$$$\sum \frac{ab}{a+b} = \sum \frac{abc}{ac+bc} =\sum \frac{1}{ac+bc} \ge \frac{9}{2(ab+bc+ac)} $$Multiplying yielts the result This question can be bashed by $Lagrange$ or $UVW$, but, nice problem

MathLuis wrote: Assassino9931 wrote: Let $a$, $b$ and $c$ be positive real numbers with $abc = 1$. Determine the minimum possible value of $$ \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) \cdot \left(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}\right) $$as well as all triples $(a,b,c)$ which attain the minimum. This one is cute By Am-Gm $$\sum_{\text{cyc}} \frac{a}{b} \ge 3$$Now by Titu's Lemma and Am-Gm $$\sum_{\text{cyc}} \frac{ab}{a+b}=\sum_{\text{cyc}} \frac{1}{\frac{1}{a}+\frac{1}{b}} \ge \frac{9}{2} \cdot \frac{1}{\sum_{\text{cyc}} \frac{1}{a}} \ge \frac{3}{2}$$And now multiplying both results we get the desired ineq with equality case: $(a,b,c)=(1,1,1)$ Thus we are done I think that in the last application $\sum \frac{1}{a} \geq 3$ by AM-GM you have an inequality in the wrong direction? More generally, if one begins with replacing the first multiplier by $3$, it's a fall into a trap if I am not mistaken!

Let $a$, $b$ and $c$ be positive numbers with $abc=1$. Prove that: $$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) \cdot \left(\frac{ab}{a+kb} + \frac{bc}{b+kc} + \frac{ca}{c+ka}\right) \geq\frac{9}{k+1}$$Where $k\geq 1.$

sqing wrote: Let $a$, $b$ and $c$ be positive numbers with $abc=1$. Prove that: $$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) \cdot \left(\frac{ab}{a+kb} + \frac{bc}{b+kc} + \frac{ca}{c+ka}\right) \geq\frac{9}{k+1}$$Where $k\geq 1.$

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