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?
 A: 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*}
A: 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.
A: Hint
The inequality is homogeneous in $(a,b,c)$ so we can choose them such that
$$a+b+c=\frac{1}{3}$$
so we have to find the minimum for the function:
$$f(a,b,c)=\frac{a}{1-2a}+\frac{b}{1-2b}+\frac{c}{1-2c}$$
with boundary
$$a+b+c=\frac{1}{3}$$
Now can use Lagrange Multiplier and find the minimum $3/7$
A: 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)
A: WLOG  $a\geq b\geq c.$  $$\text {Let } x=1/(a+3b+3c),\; y=1/(b+3c+3a),\; z=1/(c+3a+3b).$$  We have $a+3b+3c\leq b+3c+3a\leq c+3b+3a.$ Therefore $$(1).\quad x\geq y\geq z>0.$$ Let $f(a,b,c)=ax+by+cz.$
Differentiating $f$ by $c,$ keeping $a$ and $b$ constant, we have $$\partial f/\partial c=-3ax^2-3by^2+3(a+b)z^2=3(a(z^2-x^2)+b(z^2-y^2)).$$  This is not positive by (1). Therefore $$(2).\quad f(a,b,c)\geq f(a,b,b).$$ Differentiating $f$ by $a,$ keeping $b$ and $c$ constant, we have $$\partial f/\partial a=(3b+3c)x^2-3by^2-3cz^2=3(b(x^2-y^2)+c(x^2-y^2)).$$ This is not negative  by (1). Therefore $$(3). \quad f(a,b,c)\geq f(b,b,c).$$ By(2) and (3) we have $$f(a,b,c)\geq f(b,b,c)\geq f(b,b,b)=3/7.$$
A: We need to prove that 
$$\sum\limits_{cyc}\left(\frac{a}{a+3b+3c}-\frac{1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{2a-b-c}{a+3b+3c}\geq0$$ or
$$\sum\limits_{cyc}\frac{a-b-(c-a)}{a+3b+3c}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{1}{a+3b+3c}-\frac{1}{b+3c+3a}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{(a+3b+3c)(b+3c+3a)}\geq0.$$
Done!
A: Another way.
Let $a+b+c=3$.
Hence, we need to prove that
$$\sum\limits_{cyc}\frac{a}{9-2a}\geq\frac{3}{7}$$ or
$$\sum\limits_{cyc}\left(\frac{a}{9-2a}-\frac{1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{a-1}{9-2a}\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{a-1}{9-2a}-\frac{a-1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-1)^2}{9-2a}\geq0.$$
Done!
