Inequality proof verification

[methodrequired]

Prove $$\frac{a+3b}{3a+b}+\frac{b+3c}{3b+c}+\frac{c+3a}{3c+a} \geqslant 3 ~~~ , ~ \forall a,b,c >0 .$$​

$$\frac{a+3b}{3a+b}-1+\frac{b+3c}{3b+c}-1+\frac{c+3a}{3c+a}-1 \geqslant 0$$ ,​

(*)$$\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \frac{a-c}{3c+a} \geqslant 0 .$$
Assume WLOG $$a \le b \le c$$ at the denominators then we have $$\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \frac{a-c}{3c+a} \geqslant \frac{(b-a)+(c-b)+(a-c)}{4c} =0 .$$If not correct , how to proceed from (*) ?

 

[Cubist]

I think you are correct. To make it slightly easier for the other tutors to agree (or disagree) with my appraisal I'll expand some of your working...

Prove $$\frac{a+3b}{3a+b}+\frac{b+3c}{3b+c}+\frac{c+3a}{3c+a} \geqslant 3 ~~~ , ~ \forall a,b,c >0 .$$​

$$\frac{a+3b}{3a+b}-1+\frac{b+3c}{3b+c}-1+\frac{c+3a}{3c+a}-1 \geqslant 0$$ ,​

$$\left( \frac{a+3b}{3a+b}-\frac{3a+b}{3a+b} \right) + \left(\frac{b+3c}{3b+c}-\frac{3b+c}{3b+c}\right)+ \left(\frac{c+3a}{3c+a}-\frac{3c+a}{3c+a} \right)\geqslant 0$$
$$\frac{2(b-a)}{3a+b} + \frac{2(c-b)}{3b+c} + \frac{2(a-c)}{3c+a} \geqslant 0$$
divide both sides of the inequality by two...

(*)$$\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \frac{a-c}{3c+a} \geqslant 0 .$$
Assume WLOG $$a \le b \le c$$ at the denominators then we have

New statement:-

$$\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \frac{a-c}{3c+a} \geqslant \frac{b-a}{3c+c} + \frac{c-b}{3c+c} + \frac{a-c}{3c+c}$$
the above is true because each denominator increases in value (or stays the same), which has the effect of lowering the value of every term (or leaving it the same). And since...

$$\frac{(b-a)+(c-b)+(a-c)}{4c} =0 .$$

Then

$$\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \frac{a-c}{3c+a} \geqslant 0$$

 

[Cubist]

$$\frac{b-a}{3a+b} + \frac{c-b}{3b+c} + \color{red}\frac{a-c}{3c+a}\color{black} \geqslant \frac{b-a}{3c+c} + \frac{c-b}{3c+c} + \color{red}\frac{a-c}{3c+c}$$
the above is true because each denominator increases in value (or stays the same), which has the effect of lowering the value of every term (or leaving it the same). And since...

...actually I'm not sure that the above is always true since the numerators of the red terms are negative (or zero)