<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] 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 $$f(a,b,c)\geq f(b,b,c)\geq f(b,b,b)=3/7.$$
</s>