Question regarding a proof of a inequality Given $a,b,c, \in (0,\infty)$, then the following inequality holds
$$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 2 \sqrt6 (a+b+c)$$
What I've tried:
First, I noticed that we can factor
$$5a^2+12ab+7b^2 = (a+b)(5a+7b)$$
1.) Therefore we can use AM-GM with $a+b$as the first term and $(5a+7b)$ as the second term
$$\sqrt{(a+b)(5a+7b)} \leq \frac{(a+b)+(5a+7b)}{2}$$
So
$$\sqrt{5a^2+12ab+7b^2} \leq 3a+4b$$
Analogous we get
$$\sqrt{5b^2+12bc+7c^2} \leq 3b+4c$$
$$\sqrt{5c^2+12ca+7a^2} \leq 3c+4a$$
Summing all three inequalities,
$$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 7(a+b+c)$$
However, since $7>2\sqrt6$, this approach doesn't yield the wanted result.
2.) Another way was to see what constraints are on $a,b$ are if we have that
$$\sqrt{{5a^2+12ab+7b^2}} \leq \sqrt6 (a+b)$$
and analogue for $b,c$ and $c,a$.
Squaring both sides, and knowing that $a,b$ are positive real numbers, we can only consider the positive branch of the square-root,
$(a+b)(5a+7b) \leq 6(a+b)^2$ then $5a+7b \leq 6a+6b$, thus $b \leq a$.
Analogous,
$c \leq b \leq a$, but also $a \leq c$ which would lose generality because by the transitive property, $c \leq a$ and $a \leq c$ which is true if and only if $a=c$. Therefore this approach is also not correct.
3.) I don't think we can apply Minkowski's Inequality, I can't find a way to write $7b^2$ in a way to write our expression as a sum of two squares of binomials.
How can this inequality be proven? Can we use any of these ideas?

Your inequality become the equality when $a=b=c.$ But for $a=b,$ then
$$\sqrt{(a+b)(5a+7b)} \ne \frac{(a+b)+(5a+7b)}{2}.$$
By the AM-GM inequality, we have
$$\sqrt{(a+b)(5a+7b)} = \frac{1}{\sqrt6} \cdot \sqrt{6(a+b) \cdot (5a+7b)} \leqslant \frac{6(a+b)+(5a+7b)}{2\sqrt6} = \frac{11a+13b}{2\sqrt6}.$$
Equality hold for $a=b.$
Therefore
$$\sum \sqrt{(a+b)(5a+7b)} \leqslant \sum \frac{11a+13b}{2\sqrt6} = 2\sqrt 6 (a+b+c).$$
