By AM-GM, \[a+\frac{1}{b}\ge 2\sqrt\frac{a}{b}\] \[b+\frac{1}{c}\ge 2\sqrt\frac{b}{c}\] \[c+\frac{1}{a}\ge 2\sqrt\frac{c}{a}\] Multiplying the above results gives the result.

Thi is equivalent to $(ab+1)(bc+1)(ca+1) \ge 8abc$. Now, $(ab+1) \ge 2\sqrt{ab}$ and so on.Multiplication of the three inequalities gives us the desired result.

My solution:

Holder's inequality gives us that $ \left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right) \geq \left(a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} + \frac{1}{a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}}\right)^3$. Therefore, it's enough to prove that $a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} + \frac{1}{a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}} \geq 2$, what is clearly true.

it`s easy with AM-GM

$(a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{a}) = (a+\frac{1}{a})+(b+\frac{1}{b})+(c+\frac{1}{c})+(abc+\frac{1}{abc})$...(1) Now by AM-GM for $x$ and $\frac{1}{x}$: $\frac{x+\frac{1}{x}}{2}\geq \sqrt{x*\frac{1}{x}}=\sqrt{1}=1$ Thus $x+\frac{1}{x}\geq 2$ But each of the brackets of the RHS of (1) is in the form $x+\frac{1}{x}$ so (1) $\geq 4*2=8$ as required with equality if and only if $a=b=c$

Generalization. Given positive reals $a,b,c;$ show that we have\[\left(\lambda a+\frac 1b\right)\left( \mu b+\frac 1c\right)\left(\nu c+\frac 1a\right)\geq 8\sqrt{\lambda \mu \nu }.(\lambda ,\mu .\nu >0)\]

sqing wrote: Generalization. Given positive reals $a,b,c;$ show that we have\[\left(\lambda a+\frac 1b\right)\left( \mu b+\frac 1c\right)\left(\nu c+\frac 1a\right)\geq 8\sqrt{\lambda \mu \nu }.(\lambda ,\mu .\nu >0)\]

Potla wrote: Given positive reals $a,b,c;$ show that we have \[\left(a+\frac 1b\right)\left(b+\frac 1c\right)\left(c+\frac 1a\right)\geq 8.\] Given positive reals $a,b,c;$show that we have $$ (a+ \frac1a)(b+ \frac 1b)(c+ \frac 1c) \geq 8+\frac{2(a-b)^2}{ab} $$

sqing wrote: Potla wrote: Given positive reals $a,b,c;$ show that we have \[\left(a+\frac 1b\right)\left(b+\frac 1c\right)\left(c+\frac 1a\right)\geq 8.\] Given positive reals $a,b,c;$show that we have $$ (a+ \frac1a)(b+ \frac 1b)(c+ \frac 1c) \geq 8+\frac{2(a-b)^2}{ab} $$ Clearly $c+\frac{1}{c} \ge 2$ by AM-GM, thus we only need to show $$(a+\frac{1}{a})(b+\frac{1}{b}) \ge 4 + \frac{a^2-2ab+b^2}{ab} = \frac{a^2+2ab+b^2}{ab}$$Multiplying by ab, we find this is equivalent to $$a^2 b^2 + a^2 + b^2 + 1 \ge a^2 + 2ab + b^2$$$$\Leftrightarrow (ab-1)^2 \ge 0$$which is true by the trivial inequality.

AdventuringHobbit wrote: sqing wrote: Given positive reals $a,b,c;$show that we have $$ (a+ \frac1a)(b+ \frac 1b)(c+ \frac 1c) \geq 8+\frac{2(a-b)^2}{ab} $$ Clearly $c+\frac{1}{c} \ge 2$ by AM-GM, thus we only need to show $$(a+\frac{1}{a})(b+\frac{1}{b}) \ge 4 + \frac{a^2-2ab+b^2}{ab} = \frac{a^2+2ab+b^2}{ab}$$Multiplying by ab, we find this is equivalent to $$a^2 b^2 + a^2 + b^2 + 1 \ge a^2 + 2ab + b^2$$$$\Leftrightarrow (ab-1)^2 \ge 0$$which is true by the trivial inequality. Nice.Thanks. $$ (ab-1)^2 \ge 0 \Leftrightarrow (a+\frac{1}{a})(b+\frac{1}{b}) \ge 4 + \frac{(a-b)^2}{ab} $$$$ (a+ \frac1a)(b+ \frac 1b)(c+ \frac 1c) \geq 2 (a+ \frac1a)(b+ \frac 1b)\geq 8+\frac{2(a-b)^2}{ab} $$$$\implies$$Let $a,b,c $ be positive reals . Show that $$ (a+ \frac1a)(b+ \frac 1b)(c+ \frac 1c) \geq 8+\frac{2}{3} \left(\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2+\left(\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{b}}\right)^2+\left(\sqrt{\frac{c}{a}}-\sqrt{\frac{a}{c}}\right)^2 \right)$$