Not an solution, just some thoughts
Let $f(a,b,c)$ be the expression we want to minimize. Suppose one of the variables is $0$, and the other two are inverses of each other. WLOG, assume $a=0, b=\frac{1}{c}$. We get:
$f(0,\frac{1}{c},c) = \frac{\sqrt{8}}{5} + \frac{\sqrt{\frac{5}{c}}}{\frac{8}{c}} + \frac{\sqrt{5c}}{8c} = \frac{\sqrt{8}}{5} + \frac{\sqrt{5}}{8}\left(\sqrt{c} + \frac{1}{\sqrt{c}}\right)$
By AM-GM, it's now clear $f(0,\frac{1}{c},c) \geq \frac{2\sqrt{2}}{5} + \frac{\sqrt{5}}{4}$.
From this point on, assume, WLOG, $a\geq b\geq c>0$.
Claim 1: $\min(a,b,c) \leq 1$.
Proof: Suppose not. Then $ac+ab+bc=1$ is impossible. $\blacksquare$
Claim 2: If $a\geq b\geq c$, then $1 \geq b\geq c$.
Proof: Suppose not. Then $a\geq b > 1$. Meaning $ab>1$. Meaning $ab+bc+ac=1$ is impossible. $\blacksquare$
Claim 3: $abc \leq \frac{1}{3\sqrt{3}}$.
Proof: AM-GM gives $\frac{ab+bc+ac}{3} \geq \sqrt[3]{a^{2}b^{2}c^{2}} \implies \frac{1}{3} \geq \sqrt[3]{a^{2}b^{2}c^{2}} \implies \frac{1}{3\sqrt{3}} \geq abc$ $\blacksquare$
Claim 4: $a+b+c \geq \sqrt[3]{3\sqrt{3}}$
Proof: Direct application of claim 3 and AM-GM $\blacksquare$
Claim 5: $a+b+c \geq \sqrt{3}$
Let $S=a+b+c$ and $P = abc$. We can say that $a,b,c$ are the roots of an polynomial $g(x) = x^{3} - Sx^{2} + x - P$. For $g$ to have 3 real roots, we must have that $g$ has an critical point above $0$, and another below $0$. Meaning
$3x^{2} -2Sx + 1 = 0 \implies x = \frac{2S\pm2\sqrt{S^{2} - 3}}{2} = S \pm \sqrt{S^{2}-3}$
Must have real roots, so $S \geq \sqrt{3}$.$\blacksquare$
Claim 6: If $b=c$, then $a\geq \frac{1}{\sqrt{3}}$, and $b=c=\sqrt{a^{2}-1}-a$
Suppose we know that two of the variables are equal. Then, either $a=b=1$ (which we already solved), or $1 \geq b=c$. In this situation we have that $ac + ab + bc = 1 \implies b(b+2a)=1 \implies b^{2}+2ab - 1=0 \implies b=\frac{-2a\pm 2\sqrt{a^{2}+1}}{2} = \sqrt{a^{2}+1}-a$ (since $b>0$).
Enforcing $a\geq b$, we get that this has an solution whenever $a\geq \frac{1}{\sqrt{3}}$. $\blacksquare$
So, if we plug in $f(a,\sqrt{a^{2}+1}-a,\sqrt{a^{2}+1}-a)$, $a\geq \frac1{\sqrt{3}}$, we fairly easily check $f(a,\sqrt{a^{2}+1}-a,\sqrt{a^{2}+1}-a) > f(1,1,0)$
Meaning if there is an minimum, and it is not $f(1,1,0)$, it must occur with $a\neq b\neq c$. So it suffices to show this is not possible.
From this point on, we suppose $a>b>c$. We also know $1\geq b > c$.
I ran out of ideas here. I suspect one can prove that $f(a,b,c) \geq f(a,\sqrt{a^{2}+1}-a,\sqrt{a^{2}+1}-a)$ whenever $a\geq \frac{1}{\sqrt{3}}$, which would be enough, but I have been unable to do so.