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?

Note that
\begin{align*}
(a+b+c)^2 \ge 3ab+3bc+3ac.
\end{align*}
Therefore,
\begin{align*}
&\ \frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b}\\
=&\ \frac{a^2}{a^2+3ab+3ac}+\frac{b^2}{b^2+3ab+3bc}+\frac{c^2}{c^2+3ac+3bc}\\
\ge&\ \frac{(a+b+c)^2}{a^2+b^2+c^2 + 6ab + 6ac+6bc}\\
=&\ \frac{(a+b+c)^2}{(a+b+c)^2 + 4ab + 4ac+4bc}\\
\ge&\ \frac{(a+b+c)^2}{(a+b+c)^2 + \frac{4}{3}(a+b+c)^2}\\
=&\ \frac{3}{7}.
\end{align*}
