About metric space Let $d_1,d_2$ be two metrics on $X$ prove or disprove $ \sqrt{(d_1)^2 + (d_2)^2} $ is metric on $X$.
let $X= \mathbb{R} $ and  $d_1(x,y) = |x-y| $ then $(d_1(x,y))^2$ is not a metric also we know $d_1 +d_2$ is always a metric also $ \sqrt{(d_1)}$ is always metric. 
 A: There is no counterexample because $\sqrt{d_1^2+d_2^2}$ is, indeed, a metric. Note that it is obviously symmetric, and $\sqrt{d_1(x,y)^2+d_2(x,y)^2}=0$ if and only if $x=y$. Hence, we only have to check the triangle inequality.
By the triangle inequality in $\mathbb{R}^2$, we have for all $r,s,t,u\in \mathbb{R}$ that
$$\sqrt{(r+s)^2+(t+u)^2}\leq \sqrt{r^2+t^2}+\sqrt{s^2+u^2}$$
Thus, for $x,y,z\in X$, we can apply the triangle inequality for $d_1$ and $d_2$ (as well as the fact that $r\mapsto r^2$ and $r\mapsto \sqrt{r}$ are increasing functions) to get
\begin{align}
\sqrt{d_1(x,y)^2+d_2(x,y)^2} &\leq \sqrt{(d_1(x,z)+d_1(y,z))^2+(d_2(x,z)+d_2(y,z))^2}\\
&\leq \sqrt{d_1(x,z)^2+d_2(x,z)^2}+\sqrt{d_1(y,z)^2+d_2(y,z)^2},
\end{align}
Hence, $\sqrt{d_1^2+d_2^2}$ satisfies the triangle inequality and thus, it is a metric.
In general, this strategy proves that if $\|\cdot\|$ is any norm on $\mathbb{R}^n$ and $(d_j)_{1\leq j\leq n}$ are $n$ metrics on $X$, then 
$$\tilde{d}(x,y)=\| (d_j(x,y))_{1\leq j\leq n}\|$$ defines a metric on $X$ as well.
