lean_workbook_plus_47398

Squaring both sides of $\frac{c}{a}+\frac{a}{b}+\frac{b}{c}=-\frac{3}{2}$ we get $\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2}+2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)=\frac{9}{4}$ . So we are to prove $\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\geqq \frac{9}{4}-(\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2})$ . This is equivalent to $\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2}\geqq \frac{9}{4}$ , which is obvious because by AM-GM $\frac{b^2}{a^2}+\frac{a^2}{b^2} \geq 2$ , $\frac{c^2}{b^2}+\frac{b^2}{c^2} \geq 2$ and $\frac{a^2}{c^2}+\frac{c^2}{a^2} \geq 2$ and hence $\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2}\geqq 6> \frac{9}{4}$