 Hayley Bernard

2022-07-14

For $a$, $b$, $c$ the sides of a triangle, prove $|\frac{a}{b}-\frac{b}{a}+\frac{b}{c}-\frac{c}{b}+\frac{c}{a}-\frac{a}{c}|<1$
I initially thought of changing it in terms of sine using the law of sines, which gives me
$|\frac{\mathrm{sin}A}{\mathrm{sin}B}-\frac{\mathrm{sin}B}{\mathrm{sin}A}+\frac{\mathrm{sin}B}{\mathrm{sin}C}-\frac{\mathrm{sin}C}{\mathrm{sin}B}+\frac{\mathrm{sin}C}{\mathrm{sin}A}-\frac{\mathrm{sin}A}{\mathrm{sin}C}|<1$
However, from there I'm not sure how to proceed. Any tips? decoratesuw

Expert

Hint: Using the substitution $a=x+y,b=y+z,c=z+x$ (if $a,b,c$ are sides of a triangle then $x,y,z$ always exist and are positive) we get
$|\sum _{cyc}\left(\frac{a}{b}-\frac{b}{a}\right)|=|\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}|<1$ yasusar0

Expert

By an inequality of the triangle $|\sum _{cyc}\left(\frac{a}{b}-\frac{a}{c}\right)|=\frac{|\sum _{cyc}\left({a}^{2}c-{a}^{2}b\right)|}{abc}=\frac{\prod _{cyc}|a-b|}{abc}=\prod _{cyc}\frac{|a-b|}{c}<1$