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?
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Albus Dumbledore
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The trick to your question is explained in (both answers of) https://math.stackexchange.com/questions/1193504/ (a generalization of your question): consider $\frac ab + \frac ab +\frac bc$. – player3236 Dec 02 '20 at 15:25
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This is directly related to your question, and the most upvoted answer uses basically the same trick: https://math.stackexchange.com/questions/191436 – player3236 Dec 02 '20 at 15:33
2 Answers
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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.
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Hint: $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2b}{b^2c}}=3\sqrt[3]{a^3}=3a$$ $$\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\ge 3\sqrt[3]{\frac{b^2c}{c^2a}}=3\sqrt[3]{b^3}=3b$$ $$\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\ge 3\sqrt[3]{\frac{c^2a}{a^2b}}=3\sqrt[3]{c^3}=3c$$
Add all: what do you get?
Albus Dumbledore
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