<s>[INST] If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle 
$ABC$ is equilateral 
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) = \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)+3$
let $\displaystyle \frac{a}{b} = x,\frac{b}{c}=y,\frac{c}{a}=z$
so $\displaystyle 2(x+y+z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$
wan,t be able to proceed after that, could some help me
 [/INST] $x+y+z\geq 3(xyz)^{\frac{1}{3}}$,when $x=y=z$ ,equal sign is established.So:$$2\sum\frac{c}{a}\geq\sum\frac{c}{a}+3$$ i.e.$$\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\leq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$$Without loss of generality, we assume:$c\geq b\geq a$,so $\frac1{a}\geq\frac1{b}\geq\frac1{c},$then we have:$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\leq\frac{c}{a}+\frac{b}{c}+\frac{a}{b},$$so $a=b=c$.I'm using an incomplete sort to derive a contradiction.
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