<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] Cauchy-Schwarz actually works pretty well.
We have with Cauchy-Schwarz if we put $x_1=\sqrt{\frac{a}{a+3b+3c}}$, $x_2=\sqrt{\frac{b}{b+3c+3a}}$ and $x_3=\sqrt{\frac{c}{c+3a+3b}}$ as well as $y_1=\sqrt{a(a+3b+3c)}$, $y_2=\sqrt{b(b+3c+3a)}$ and $y_3=\sqrt{c(c+3a+3b)}$
$$
\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)≥\left(x_1y_1+x_2y_2+x_3y_3\right)^2\iff\\
\left(\frac{a}{a+3b+3c}+\frac{b}{b+3c+3a}+\frac{c}{c+3a+3b}\right)\left(a(a+3b+3c)+b(b+3c+3a)+c(c+3a+3b)\right)≥(a+b+c)^2
$$
so its enough to prove
$$
\frac{(a+b+c)^2}{\sum_{cyc}a(a+3b+3c)}≥\frac37\iff\\
7(a^2+b^2+c^2+2ab+2bc+2ca)≥3(a^2+b^2+c^2+6ab+6bc+6ca)\iff\\
a^2+b^2+c^2≥ab+bc+ca
$$
which is true.
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