$\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geq\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}$ for any real numbers. I think that this inequality is strong, though I do not have knowledge of many techniques. There goes my work:
Positive variables only make the inequality stronger, hence suppose $a,b,c\geqslant0$
$$
\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geqslant\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}
$$By squaring,
$$
\Rightarrow
\sqrt{(a^2+5b^2)(b^2+5c^2)}+\sqrt{(b^2+5c^2)(c^2+5a^2)}+\sqrt{(c^2+5a^2)(a^2+5b^2)}\geq2(a+b+c)^2
$$The $LHS$
$$=
\sqrt{\sum_{cyc}{5b^4 + 31a^2b^2 + 2\left(a^2 + 5b^2\right) \left(\sqrt{\left(b^2 + 5c^2\right) \left(c^2 + 5a^2\right)}\right)}}
$$$$
\geqslant 
\sqrt{\sum_{cyc}{5b^4 + 31a^2b^2 + 2(a^2 + 5b^2)(bc + 5ca)}}
$$
Now we are only left to prove that
$$
\sum_{cyc}{5b^4 + 31a^2b^2 + 52a^2bc + 10a^3c + 10a^3c} \geqslant \sum_{cyc}{4a^4 + 16(a^3b + ab^3) + 24a^2b^2 + 48a^2bc}
$$$$
\sum_{cyc}{a^4 + 7a^2b^2 + 4a^2bc - 6(a^3b + ab^3)} \geqslant 0
$$
The last inequality is wrong for $(a,b,c) = (1,1,0)$. Cauchy Schwarz looks fine but I am not able to find a way.
I found this inequality posted by arqady on aops forum.
Please help!

Probably not the proof you are looking for, but a proof nonetheless.
The inequality is really sharp, and I don't think that a manual solution exists. Concretely, I don't think that one can find a lower bound on the LHS, such that we can algebraically confirm that it upper bounds the RHS. However, it is easy to numerically verify that the inequality holds, and I hope that you can find this convincing.
Specifically, divide both sides by $\sqrt{a^2 + b^2 + c^2}$, then we're left with the equivalent inequality:
$$
\sqrt{x^2 + 5y^2} + \sqrt{y^2 + 5z^2} + \sqrt{z^2 + 5x^2} \geq \sqrt{10 + 8(xy + yz + xz)},
$$
where $x = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, y = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, z = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$, and $x^2 + y^2 + z^2 = 1$. Furthermore, it has been established that we can safely assume that $x,y,z\geq 0$, so it is sufficient to verify the inequality on the surface $\{(x,y,z) \in\mathbb{R}^3 ~\vert~ x^2 + y^2 + z^2 = 1, x,y,z\geq 0\}$, which can be parameterized with $$x = \sin\theta\sin\omega,\quad y = \sin\theta\cos\omega,\quad z=\cos\theta,$$ with $(\theta,\omega)\in[0,\pi/2]\times[0,\pi/2]$.
Now, if one minimizes the function
$$
h(\theta,\omega) = \sqrt{x^2 + 5y^2} + \sqrt{y^2 + 5z^2} + \sqrt{z^2 + 5x^2} - \sqrt{10 + 8(xy + yz + xz)},
$$
over the square $[0,\pi/2]\times[0,\pi/2]$, one then finds that it has a unique global minimum 0 at $x=y=z=\frac{1}{\sqrt{3}}$, or at $\theta \approx 0.9554,~ \omega = \pi/4$, see the figure below which shows the level sets of $h$.
$h(\theta,\omega)$" />
This implies by homogeneity that the original inequality is equality only at $a=b=c$, and a strict inequality at all other values.
