How to prove $\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac b{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}$ when $a,b,c>0$ I want to prove that for $a,b,c>0$ we have
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=
\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}.$$
My first attempt: By Cauchy-Schwarz we have $$\left(\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}\right)^2\le3\sum_{cyc}\frac{a^2}{a^2+3b^2+3c^2}$$ so we only need to prove that the right-hand side is always less than $\frac{9}{7}$, but this is false. Failed
Second attempt: By Cauchy-Schwarz
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=\sum_{cyc} \frac 1{\sqrt{1+3\frac{b^2}{a^2}+3\frac{c^2}{a^2}}}\le\sum_{cyc} \frac{\sqrt 7}{1+3\frac{b}{a}+3\frac{c}a}$$
so it remains to prove that $$\sum_{cyc} \frac{a}{1+3b+3c}\le\frac37$$ but this is wrong for example for $a=1,b=1,c=2$. Failed
Third attempt: Let $S=3(a^2+b^2+c^2)$. We need to prove $$\sum_{cyc} \frac{a}{\sqrt{S-2a^2}}\le \frac37.$$ But $x\mapsto \frac{x}{\sqrt{S
-2x^2}}$ is convex so Jensen has the wrong direction...

We need to prove that:
$$\sqrt{\frac{a}{a+3b+3c}}+\sqrt{\frac{b}{b+3c+3a}}+\sqrt{\frac{c}{c+3a+3b}}\leq\frac{3}{\sqrt7}$$
Let's normalize with $a+b+c=3$. Then the inequality is equivalent with:
$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}}+\sqrt{\frac{c}{9-2c}}\leq \frac{3}{\sqrt{2}}$$
Without loss of generality suppose that $a\le b\le c$. Then we have $a+b\leq 2$ and we will prove:
$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}} \leq \sqrt{\frac{2(a+b)}{9-a-b}}$$
Squaring twice, this is equivalent with:
$$\frac{(a-b)^2[729+81(a+b)^2-486(a+b)-16ab(a+b)]}{(9-2a)^2(9-2b)^2(9-a-b)^2}\geq 0$$
We have $16ab(a+b)\leq 16(a+b)$ and 
$$729+81x^2-502x\geq 0, \text{ when }x \leq 2$$ 
It remains to prove that:
$$\sqrt{\frac{14(3-c)}{6+c}}+\sqrt{\frac{7c}{9-2c}}\leq 3$$
Squaring twice this is equivalent with:
$$\frac{81(c-1)^2 (12 - 5 c)^2}{(9 - 2 c)^2 (6 + c)^2}\geq 0$$
which gives the two equality cases.
