The longer

This won't be a thorough solution, but rather of rundown of the steps needed to do to be able to solve the problem. Other solutions are very welcome :)

Let us take note that $\displaystyle \int_0^{\infty} e^{-ax} \mathrm{d}x = \frac{1}{a}$ , right?

From this, we can infer that the integrand could be interpreted as the definite integral of a dummy variable $t$ , such as

$\frac{3}{x+y+z} = \displaystyle\int_0^{\infty} e^{ - \frac{x+y+z}{3}t} \mathrm{d}t$

This now changes the integral to

$\displaystyle \int_0^1 \int_0^1 \int_0^1 \! \dfrac{3}{x+y+z} \:\: \mathrm{d}x \mathrm{d}y \mathrm{d}z = \int_0^1 \int_0^1 \int_0^1 \! \mathrm{d}x \mathrm{d}y \mathrm{d}z \int_0^{\infty} \! e^{ - \frac{x+y+z}{3}t} \: \mathrm{d}t$

We can now change the order of integration to simplify this to

$\displaystyle \int_0^{\infty} (\int_0^1 e^{- \frac{t}{3}x} \mathrm{d}x)^3 \mathrm{d}t$

$\displaystyle \int_0^{\infty} (\frac{1 - e^{- \frac{t}{3}}}{\frac{t}{3}})^3 \mathrm{d}t$

$\displaystyle \int_0^{\infty} (\frac{3 (1-e^{-\frac{t}{3}})}{t})^3 \mathrm{d}t$

$\displaystyle \int_0^{\infty} \frac{27}{t^3} - \frac{81e^{-t/3}}{t^3} + \frac{81e^{-2t/3}}{t^3} - \frac{27e^{-t}}{t^3} \mathrm{d}t$

This will ultimately give us $\frac{9}{2} \ln \frac {27}{16}$ , such that $9+2+27+16 = 54$ .

Rather than struggling with the singularities (the details behind the word "ultimately") in Efren's proof, it is easier to do this directly: $\begin{aligned} \int_0^1 \frac{3}{x+y+z}\,dz & = \Big[3\ln(x+y+z)\Big]_{z=0}^1 \; = \; 3\ln(x+y+1) - 3\ln(x+y) \\ \int_0^1\left(\int_0^1 \frac{3}{x+y+z}\,dz\right)\,dy & = \int_0^1 \big[3\ln(x+y+1) - 3\ln(x+y)\big]\,dy \\ & = 3\Big[(x+y+1)\ln(x+y+1) - (x+y+1) - (x+y)\ln(x+y) + (x+y)\Big]_{y=0}^1 \\ & = 3(x+2)\ln(x+2) - 6(x+1)\ln(x+1) + 3x\ln x \\ \int_0^1 \int_0^1 \int_0^1 \frac{3}{x+y+z}\,dx\,dy\,dz & = 3\int_0^1 \big[(x+2)\ln(x+2) - 2(x+1)\ln(x+1) + x\ln x\big]\,dx \\ & = 3\Big[ \tfrac12(x+2)^2\ln(x+2) - \tfrac14(x+2)^2 - (x+1)^2\ln(x+1) + \tfrac12(x+1)^2 + \tfrac12x^2\ln x - \tfrac14x^2\Big]_0^1 \\ & = 3\Big(\big(\tfrac92\ln3 - \tfrac94 - 4\ln2 + 2 - \tfrac14\big) - \big(2\ln2 - 1 + \tfrac12\big)\Big) \; = \; 3\big(\tfrac92\ln3 -4\ln2 - 2\ln2\big) \\ & = \tfrac92\ln\tfrac{27}{16} \end{aligned}$