Solution: Recall that $d$ is a metric on a space $X$ if for all $x,y,z\in X$ we have

• $d(x,y) \geq 0$
• $d(x,y) = 0 \iff x=y$
• $d(x,y) = d(y,x)$ and
• $d(x,y) + d(y,z) \leq d(x,z)$.

1. Given that $d(x,y) = d_1(x,y) + d_2(x,y)$. Let $x,y,z\in X$. We have
   • Since $d_1(x,y) \geq 0$ and $d_2(x,)\geq 0$ so, $d(x,y) = d_1(x,y) + d_2(x,y) \geq 0$.
   • If $d(x,y) = 0$ then $d_1(x,y) + d_2(x,y) = 0$ which implies $d_1(x,y) = d_2(x,y) = 0$ as both are non-negative. Therefore, $x=y$.
   • Since $d_1$ and $d_2$ are symmetric $d$ will also be symmetric, that is, $d(x,y) = d_1(x,y) + d_2(x,y) = d_1(y,x) + d_2(y,x) = d(y,x)$.
   • Finally, the triangle inequality, \begin{align*} d(x,y) + d(y,z) & = d_1(x,y) + d_2(x,y) + d_1(y,z) + d_2(y,z) \\ & = d_1(x,y) + d_1(y,z) + d_2(x,y) + d_2(y,z) \\ & \leq d_1(x,z) + d_2(x,z) \\ & = d(x,z). \end{align*}

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2. This is also similar to the previous one. Let $x,y,z\in X$. We have
   • Since $d_1(x,y), d_2(x,y) \geq 0$, so $\rho(x,y) = \max \left\{ d_1(x,y),d_2(x,y) \right\} \geq 0$.
   • If $\rho(x,y) = 0$, then $\max\left\{ d_1(x,y) , d_2(x,y) \right\} = 0 $. Since both quantities are non-negative, $d_1(x,y)=0 = d_2(x,y)$. Therefore, $x=y$.
   • Since $d_1$ and $d_2$ are symmetric $\rho$ will also be symmetric, that is, $\rho(x,y) = \max\left\{ d_1(x,y) , d_2(x,y) \right\} = \max \left\{ d_1(y,x) , d_2(y,x) \right\} = \rho(y,x)$.
   • Finally, the triangle inequality, \begin{align*} \rho(x,y) + \rho(y,z) & = \sum_{i=1}^2\max\left\{ d_i(x,y),d_i(x,y) \right\} \\ & = \max\left\{ d_1(x,y)+ d_1(y,z),d_2(x,y)+d_2(y,z) \right\} \\ & \leq \max\left\{ d_1(x,z),d_2(x,z) \right\}\\ & = \rho(x,z). \end{align*}