methodrequired
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- Feb 20, 2022
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Prove [math]\frac{a+3b}{3a+b}+\frac{b+3c}{3b+c}+\frac{c+3a}{3c+a} \geqslant 3 ~~~ , ~ \forall a,b,c >0 .[/math]
[math]\frac{a+3b}{3a+b}-1+\frac{b+3c}{3b+c}-1+\frac{c+3a}{3c+a}-1 \geqslant 0[/math] ,
(*)[math]\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \frac{a-c}{3c+a} \geqslant 0 .[/math]
Assume WLOG [math] a \le b \le c [/math] at the denominators then we have [math]\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \frac{a-c}{3c+a} \geqslant \frac{(b-a)+(c-b)+(a-c)}{4c} =0 .[/math]If not correct , how to proceed from (*) ?