shouldn't inequalities be posted here?

Yes, I posted in the wrong spot! My apologies! Could someone be so kind as to move this topic to the inequalities section? Thanks!

It's easy from Am-Gm and the proposed inequality for $ a,b,c >0 $ such that $ abc \le 1 $.

let a=x/y,b=y/z,c=z/x ... and AM-GM kills it

Or you can just \[\frac ab + \frac ab+ \frac bc \ge 3 \sqrt[3]{\frac{a^2}{bc}} = 3a \]

Another solution: $ \frac{ab+bc+ca}{abc}\ge{a+b+c} $ $ ab+bc+ca\ge{a+b+c} $ Homogenize multiplying $ RHS $ by $ \sqrt[3]{abc} $ we get $ \sum{\sqrt[3]{a^3b^3}}\ge{\sum{\sqrt[3]{a^4bc}}} $ take $ a=\sqrt[3]{x} $ ; $ b=\sqrt[3]{y} $ ; $ c=\sqrt[3]{z} $ we get $ \sum{a^3b^3}\ge{\sum{a^4bc}} $ Which is just rearrangement.

GoJensenOrGoHome wrote: $ \sum{a^3b^3}\ge{\sum{a^4bc}} $ . This is false ! Actually by rearrangement, $ \sum{a^3b^3}\le{\sum{a^4bc}} $

Oh sorry my bad =(

niraekjs wrote: Prove that if $a,b,c$ are positive numbers with $abc=1$, then \[\frac{a}{b} +\frac{b}{c} + \frac{c}{a} \ge a + b + c. \] $\frac{a}{b}+\frac{2c}{a}=3c+\frac{(a-1)^2(a+2)}{a^2b}\ge 3c$ $ ...$

Prove that if $a,b,c$ are positive numbers with $abc=1$, then $$\frac{a}{b} +\frac{b}{c} + \frac{c}{a} \ge a+b+c \geq \frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1}.$$here

This is stronger: $ a,b,c>0,abc=1\Rightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge a+b+c+ab+bc+ca $ See EXCALIBUR... ____ Marin Sandu

From the condition we have that $c=\frac{1}{ab}$ ...(1) After multiplying everything and using $(1)$ we get: $a^3+a^3b^3+1 \geq a^3b+a^2b^2+a$ $...(*)$ But this is true because from $AM-GM$ we have: $a^3+1+1 \geq \sqrt[3]{a^3*1*1}=3a$ $a^3b^3+a^3b^3+1 \geq \sqrt[3]{a^3b^3*a^3b^3*1}=3a^2b^2$ $a^3+a^3+a^3b^3\geq \sqrt[3]{a^3*a^3*a^3b^3}=3a^3b$ We add these all and we subtract them all by $3$ then we get $(*)$