Can you calculate that?

Note that $a,b,c$ are the roots of the equation $f(X) = 0$ , where $f(X) \; = \; (X-a)(X-b)(X-c) \; = \; X^3 - uX^2 + (4-v)X - v$ for some constants $u,v$ . But then $\begin{aligned}(2+a)^{-1} + (2+b)^{-1} + (2+c)^{-1} & = \; \frac{(2+b)(2+c) + (2+a)(2+c) + (2+a)(2+b)}{(2+a)(2+b)(2+c)} \\ & = \; -\frac{f'(-2)}{f(-2)} \; = \; -\frac{12 + 4u + 4-v}{-8-4u-8 + 2v-v} \; = \; -\frac{16+4u-v}{-16-4u+v} \; = \; \boxed{1} \end{aligned}$

After reviewing the above solutions, I give an unnatural and tricky explanation here:

Because: $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}-1=-\frac{ab+ba+ca+abc-4}{\left ( a+2 \right )\left ( b+2 \right )\left ( c+2 \right )}=0,$

So, $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1.$

Solve $a\, b\, c+a\, b+a\, c+b\, c=4$ for $a$ give $a\to \frac{4-b c}{b c+b+c}$ .

Substitute into $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$ that value for $a$ gives $\frac{1}{\frac{4-b c}{b c+b+c}+2}+\frac{1}{b+2}+\frac{1}{c+2}$ .

Combining the terms into a single fraction gives $1$ .

The restriction to positive reals is not necessary. It works even with $\mathbb{C}$ . One does need to avoid a, b or c equaling -2.

Clearly, the given equation is satisfied by a = b = c = 1, by inspection. Ergo, 1/(a + 2) + 1/(b + 2) + 1/(c + 2) = 1/3 + 1/3 + 1/3 = 1.

Let $S_n$ denote the $n^\text{th}$ symmetric sum of $a,b,c$ . Denote $S_1 = u, S_2 = v$ . By Vieta's formula, the polynomial $f(x) := x^3 - ux^2 + vx - (4-v)$ has roots $a,b,c$ . Similarly, $f(x-2) = 0$ has roots $a + 2, b+2, c+2$ . Since $g(x):= f(x-2) = x^3 + x^2 ( -u - 6) + x(12+ 4u + v) + (-4u - v - 12),$ then we know that $x^3 \cdot g\left(\frac 1x \right) = 0$ has roots $\frac1{a+2} , \frac1{b+2} , \frac1{c+2}$ . $x^3 \cdot g\left(\frac 1x \right) = (-4u - v - 12)x^3 + x^2 (12+ 4u + v) + x (-u - 6) + 1$ By Vieta's, the expression in question is equal to $- \frac{ 12+ 4u + v}{-4u - v - 12} = \boxed1$ .

Antisolution: We know that $a=b=c=1$ satisfy the given constraint. Substitute these values into the expression in question, and we have our answer!