Prove the inequality contains reciprocal fraction I would be grateful for any tips or solution for this fractional inequality!

$(\frac{a}{b})^2+(\frac{b}{c})^2+(\frac{c}{a})^2\geq\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$

You may have already found the solution with the Stiftung Warentest hint. Its hint is interesting because it is a form that we know very well!
Suppose $a, b, c \ne 0$ and let $x = \frac{a}{b}$, $y = \frac{b}{c}$ and $z = \frac{c}{a}$.
Then we get $x^2 + y^2 + z^2 = xy + xz + yz$.
Multiplying by $2$ and arranging the terms, we have $(x^2 - 2xy +y^2) + (y^2 - 2yz + z^2) + (z^2 - 2xz + x^2) \geq 0$.
This is equivalent to $(x - y)^2 + (y - z)^2 + (z-x)^2 \geq 0$, which is true because every square is greater than $0$.
I hope this solution can help you.
