It's Sunday

Call the expression P

By observation, we see that $f(1;1;0)=\frac{5}{2}$ . So now we'll have to prove $P\geq \frac{5}{2}$ Without losing generality, assuming $c\leq b\leq a$ , now we choose a number $t>0$ satisfy $a\geq t\geq b\geq c$ and $t^2+2tc=ab+bc+ca=1 \Leftrightarrow (t+c)^2=(a+c)(b+c)=1+c^2$

Now we prove $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\geq \frac{2}{t+c}+\frac{1}{2t}$ $\Leftrightarrow\bigg(\frac{1}{\sqrt{a+c}}-\frac{1}{\sqrt{b+c}}\bigg)^2\geq\frac{(\sqrt{a+c}-\sqrt{b+c})^2}{2t(a+b)}$ $\Leftrightarrow (a+c)(b+c)\leq 2t(a+b)$ which is right since $a\geq t\geq b\geq c$

So now we need to prove $\frac{2}{t+c}+\frac{1}{2t}\geq\frac{5}{2}$ when $a=b\geq c$ .

From the condition of t we have $c=\frac{1-t^2}{2t}\geq 0$ so we also find out that $t\leq 1$

Subtitute and we get $\frac{4t}{t^2+1}+\frac{1}{2t}\geq\frac{5}{2}$ $\Leftrightarrow \frac{-5t^3+9t^2-5t+1}{2t(t^2+1)}\geq 0$ Clearly see that the denominator is bigger than 0 so all we have left is $-5t^3+9t^2-5t+1\geq 0$ $\Leftrightarrow (1-t)(5t^2-4t+1)\geq 0$ The above inequality is true because $t\leq 1$ and $5t^2-4t+1>0$ so the equality hold when $t=1\Rightarrow (a;b;c)=(1;1;0)$ and its permutations

$\color{#D61F06}{\text{Upon review, I found this solution wrong. I have failed to prove the minimum for positive reals.}}$

$\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}$

$=\frac{c}{ac+bc}+\frac{b}{ab+bc}+\frac{a}{ab+ac}$

$=\frac{c}{1-ab}+\frac{b}{1-ac}+\frac{a}{1-bc}$

$\geq \frac{{(\sqrt{a}+\sqrt{b}+\sqrt{c})}^2}{3-(ab+ac+bc)}=\frac{{(\sqrt{a}+\sqrt{b}+\sqrt{c})}^2}{2}$ by Cauchy-Schwarz inequality (T2' lemma)

$\geq \frac{9}{2}{(abc)}^{\frac{1}{3}}$ by AM-GM inequality, with equality only holds when $a=b=c=\frac{1}{\sqrt{3}}$

Therefore, if $a,b,c$ are all positive reals, the minimum is achived at $\frac{9}{2}({\frac{1}{\sqrt{3}}})=\frac{3\sqrt{3}}{2}>\frac{5}{2}$ .

Now consider one of $a,b$ or $c$ to be $0$ . Assuming $a=0$ , the expression becomes $\frac{1}{b}+\frac{1}{c}+\frac{1}{b+c}$ and we obtain $b=\frac{1}{c}$ from the constraint.

Writing everything in terms of $b$ , we get:

$b+\frac{1}{b}+\frac{1}{b+\frac{1}{b}}$ .

Now, applying the AM-GM inequality would yield a minimum of $2$ . That sounds fantastic, until I realize that this requires $b+\frac{1}{b}=1$ , for which there is no real solution. Boo, what a letdown!

Now that the simple approach won't work, a little bashing using calculus seems like a nice idea...

Write $f(b)=b+\frac{1}{b}+\frac{1}{b+\frac{1}{b}}=\frac{{(b^{2}+1)}^{2}+b^{2}}{b(b^{2}+1)}$ . Compute the first derivative as $f'(b)=\frac{b^{6}-1}{b^{2}{(b^{2}+1)}^{2}}$ . Setting $f'(b)=0$ , we find that there is only one turning point when $b=1$ , lucky us! Computing the second derivative gives $f''(1)=\frac{3}{2}$ , so the point is indeed a minimum point and we can rest easy. Doing the same thing to $f(c)$ yields the same result, so minimum is achieved when $b=c=1$ . We check that the constraint $b=\frac{1}{c}$ is satisfied.

Therefore, when not all $a,b,c$ are positive, the minimum is achieved when $a=0,b=1,c=1$ or the its permutation. This gives a minimum of $\boxed{\frac{5}{2}}$ .