Inequality using AM GM

How could you prove using only the AM GM inequality:

$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge a + b + c$$

Where $$abc = 1$$ and $$a,b,c$$ are positive real numbers?

• The trick to your question is explained in (both answers of) math.stackexchange.com/questions/1193504 (a generalization of your question): consider $\frac ab + \frac ab +\frac bc$. Dec 2, 2020 at 15:25
• This is directly related to your question, and the most upvoted answer uses basically the same trick: math.stackexchange.com/questions/191436 Dec 2, 2020 at 15:33

Hint: $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2b}{b^2c}}=3\sqrt[3]{a^3}=3a$$ $$\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\ge 3\sqrt[3]{\frac{b^2c}{c^2a}}=3\sqrt[3]{b^3}=3b$$ $$\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\ge 3\sqrt[3]{\frac{c^2a}{a^2b}}=3\sqrt[3]{c^3}=3c$$
Because $$abc=1$$ we can consider some real numbers $$x,y,z$$ such that $$a=\frac{x}{y}$$, $$b=\frac{y}{z}$$, $$c=\frac{z}{x}$$. Computing this into your equation, we get:
$$\frac{\frac{x}{y}}{\frac{y}{z}}+\frac{\frac{y}{z}}{\frac{z}{x}}+\frac{\frac{z}{x}}{\frac{x}{y}}\geq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$ which is equivalent to $$\frac{xz}{y^2}+\frac{yx}{z^2}+\frac{zy}{x^2}\geq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$ and multiplying by $$(xyz)^2$$ both sides, we get $$(xy)^3+(yz)^3+(zx)^3\geq x^3z^2y+y^3x^2z+z^3y^2x$$
Now write $$xy=m$$, $$yz=n$$, $$zx=p$$ and we get $$m^3+n^3+p^3\geq p^2m+n^2m+m^2p$$ which is a known inequality.