To continue Whubers answer, based on independence assumption between the random variables $X, Y, Z$.
$$\mathbb{E}[X\mathbb{I}_{X<Y}\mathbb{I}_{X<Z}] = \int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}x \mathbb{I}_{x<y} \mathbb{I}_{x<z}f_{X}(x)f_{Y}(y)f_{Z}(z)dzdydx$$
$$=\int_{0}^{\infty}\int_{0}^{\infty}x\mathbb{I}_{x<y}f_{X}(x)f_{Y}(y)\int_{x}^{\infty}f_{Z}(z)dz \ dydx$$
For exponential distributions with parameter $\lambda_{z}$ it holds that $F(Z>x) = \int_{x}^{\infty}f_{Z}(z)dz = e^{-\lambda_{z}x}$
$$=\int_{0}^{\infty}\int_{0}^{\infty}x\mathbb{I}_{x<y}f_{X}(x)f_{Y}(y) e^{-\lambda_{z}x} \ dydx$$
$$= \int_{0}^{\infty}xe^{-\lambda_{z}x}f_{X}(x) \int_{x}^{\infty}f_{Y}(y)dy \ dx$$ $$ = \int_{0}^{\infty}xe^{-(\lambda_{z}+\lambda_{y})x} f_{X}(x)dx$$
Now using Integration by parts we have
$$= -\frac{\lambda_{x}}{\lambda_{z}+\lambda_{y}+\lambda_{x}}\int_{0}^{\infty}x(e^{-(\lambda_{z}+\lambda_{y}+\lambda_{x})x})^{'}dx = -\frac{\lambda_{x}}{\lambda_{z}+\lambda_{y}+\lambda_{x}} (0 - \int_{0}^{\infty}e^{-(\lambda_{z}+\lambda_{y}+\lambda_{x})x}dx ) = (\frac{1}{\lambda_{z}+\lambda_{y}})^{2}$$$$= -\frac{\lambda_{x}}{\lambda_{z}+\lambda_{y}+\lambda_{x}}\int_{0}^{\infty}x(e^{-(\lambda_{z}+\lambda_{y}+\lambda_{x})x})^{'}dx = -\frac{\lambda_{x}}{\lambda_{z}+\lambda_{y}+\lambda_{x}} (0 - \int_{0}^{\infty}e^{-(\lambda_{z}+\lambda_{y}+\lambda_{x})x}dx ) = \frac{\lambda_{x}}{(\lambda_{z}+\lambda_{y}+\lambda_{x})^{2}}$$