I recently picked up a BMO 2 question from 2005:
- Let $a$, $b$, $c$ be positive real numbers. Prove that
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2 ≥ \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
However, I wasn't sure if the method I found to prove this inequality is valid although I can't see any flaws in it. Even so, I wondered if there is a specific name given to the logic used in this kind of proof, and would be interested in seeing other proofs which use a similar method.
Proof:
Let $a, b, c \in \mathbb{R}^+$.
By the AM-GM:
$${\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\over3} ≥ \sqrt[3]{\left(\frac{a^2}{b^2}\right)\left(\frac{b^2}{c^2}\right)\left(\frac{c^2}{a^2}\right)}$$
$\implies$
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} ≥ 3$$
Also, wlog (by switching variables $b$ and $c$):
$$\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2} ≥ 3$$
At least one of these 2 inequalities must be true:
$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c} ≥ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
and/or $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} ≥ \frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
$\huge\iff$
$$3+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} ≥ 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
and/or $$3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} ≥ 3+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
$\huge\implies$
(using the above inequalities)
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} ≥ 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
and/or $$\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} ≥ 3+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
$\huge\iff$
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{2b}{a}+\frac{2c}{b}+\frac{2a}{c} ≥ 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
and/or $$\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{2a}{b}+\frac{2b}{c}+\frac{2c}{a} ≥ 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
$\huge\iff$
(factorising)
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2 ≥ \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
and/or $$\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2 ≥ \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
However, by switching variables $b$ and $c$, we can see that:
$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2 ≥ \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \iff \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2 ≥ \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$
Therefore, both $$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2 ≥ \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
and $$\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2 ≥ \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
are true which proves the original inequality $_\square$
As far as I can see, all of the steps are valid. The proof requires the use of two statements of which at least one is true. The closest thing I can think of which requires this kind of logic is solving Sudokus (like the colouring method, for example).
However, the extraordinary thing is I seem to end up finding out that these two statements imply each other — either both are true or both are not. This seems to be a counterintuitive conclusion which makes me slightly unsure if this is quite right.
