<s>[INST] 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...
 [/INST] 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!
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