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...
 A: We need to prove that:
$$\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\leq\frac{3}{\sqrt7},$$ where $a$, $b$ and $c$ are positive numbers.
Indeed, by C-S $$\left(\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\right)^2\leq\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\sum_{cyc}(17a+2b+2c).$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\leq\frac{3}{49(a+b+c)}.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\sum_{cyc}\frac{a}{(9u-2a)(6u+15a)}\leq\frac{1}{49u}$$ or
$$49u\sum_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)\leq3\prod_{cyc}((9u-2a)(2u+5a)).$$
We'll prove that the last inequality is true even for any reals $a$, $b$ and $c$.
Indeed, since $\sum\limits_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)$ is a fifth degree polynomial, 
the last inequality is equivalent to $f(w^3)\geq0,$ where
$$f(w^3)=-3000w^6+A(u,v^2)w^3+B(u,v^2).$$
But $f$ is a concave function.
Thus, $f$ gets a minimal value for an extreme value of $w^3$, which happens for equality case of two variables.
Since the last inequality is an even degree, homogeneous and symmetrical, it's enough to assume $b=c=1,$ which gives $$\frac{a}{(a+6)(17a+4)}+\frac{2}{(3a+4)(2a+19)}\leq\frac{3}{49(a+2)}$$ or $$(a-1)^2(a-8)^2\geq0$$ and we are done!
A: Alternative proof:
Use the argument in https://artofproblemsolving.com/community/c6h1822770p12198977  or https://artofproblemsolving.com/community/c6h548438p3180154.
Let $x, y, z > 0$ such that
$\frac{7a^2}{a^2 + 3b^2 + 3c^2} = x^2, \
\frac{7b^2}{b^2 + 3c^2 + 3a^2} = y^2, \
\frac{7c^2}{c^2 + 3a^2 + 3b^2} = z^2$.
It is not hard to obtain
$$F(x, y, z) = 4 x^2 y^2 z^2+8 x^2 y^2+8 x^2 z^2+8 y^2 z^2+7 x^2+7 y^2+7 z^2-49  = 0.$$
It suffices to prove that if $x, y, z > 0$ and $F(x,y,z) = 0$, then $x+y+z \le 3$.
It suffices to prove that if $x, y, z > 0$ and $x + y + z > 3$, then $F(x, y, z) > 0$.
Note that $F(\alpha x, \alpha y, \alpha z) > F(x, y, z)$ for any $\alpha > 1$
and $x, y, z > 0$.
Thus, it suffices to prove that if $x, y, z > 0$ and $x+y+z = 3$,
then $F(x, y, z) \ge 0$. 
We use pqr method. Let $p = x + y + z = 3$, $q = xy + yz + zx$ and $r = xyz$. 
From $p^2 \ge 3q$, we have $q \le 3$. Let $q = 3(1-u^2)$ for $0 \le u\le 1$.
We have
\begin{align}
(x-y)^2(y-z)^2(z-x)^2 &= -4p^3r+p^2q^2+18pqr-4q^3-27r^2\\
& = 108u^6 - 27(3u^2+r-1)^2
\end{align}
which results in $108u^6 - 27(3u^2+r-1)^2 \ge 0$ and hence $r \le (2u+1)(1-u)^2 \le 3$.
Thus, we have
\begin{align}
F(x, y, z) &= 7p^2-16pr+8q^2+4r^2-14q-49 \\
&= 4(6-r)^2+72u^4-102u^2-100\\
&\ge 4(6 - (2u+1)(1-u)^2)^2 + 72u^4-102u^2-100\\
&= 2u^2(2u^2-4u+9)(2u-1)^2\\
&\ge 0.
\end{align}
We are done.
A: 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.
