<s>[INST] Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all  $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?
 [/INST] moving $$\frac{3}{7}$$ to the left and clearing the denominators we get
$$3\,{a}^{3}-{a}^{2}b-{a}^{2}c-a{b}^{2}-3\,abc-a{c}^{2}+3\,{b}^{3}-{b}^{
2}c-b{c}^{2}+3\,{c}^{3}
\geq 0$$ and this is equivalent to $$a^3+b^3+c^3-3abc+(a-b)(a^2-b^2)+(b^2-c^2)(b-c)+(a^2-c^2)(a-c)\geq 0$$
which is clear
(the first Terms are AM-GM)
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