Given two metrics $\rho,\sigma$ on $\chi$ show that the following is a metric on $\chi$ The problem asks us to show that
$$
\rho_{2}=(\rho^2+\sigma^2)^{1/2}
$$
is also a metric on $\chi$.
I'm having trouble showing the triangle inequality; I've tried numerous algebraic manipulations (filling white boards) but haven't gotten it to work. I thought I could apply the Cauchy-Schwarz inequality but these metrics aren't necessarily on $\mathbb{R}^{n}$.
I feel I am missing something obvious. 
Any hints or proofs would be appreciated. Thank you.
 A: Let $x,y,u,v \ge 0$, we first want to prove that 

$$\sqrt{(x + y)^2 + (u + v)^2} \le \sqrt{x^2 + u^2} + \sqrt{y^2 + v^2}. \tag 1$$

Squaring both sides and simplifying, the previous inequality is equivalent to $$xy + uv \le \sqrt{(x^2 + u^2)(y^2 + v^2)}.$$
Squaring again we obtain 
\begin{align}
x^2y^2 + u^2v^2 + 2xyuv \le &\ x^2y^2 + x^2v^2 + u^2y^2 + u^2v^2 \\
\iff 2xyuv \le &\ x^2v^2 + u^2y^2 \\
\iff 0 \le &\ x^2v^2 + u^2y^2 - 2xyuv \\
\iff 0 \le &\ (xv - uy)^2.
\end{align}
This proves $(1)$.

Notice also that $\frac{\partial }{\partial x}\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}} \ge 0$ for $x \ge 0$. This means that $x \mapsto \sqrt{x^2 + y^2}$ is an increasing function on $[0,\infty)$. The same holds for $y \mapsto \sqrt{x^2 + y^2}$.

Let's put everything together: 
\begin{align}
\rho_2(x,y) \le &\ \sqrt{[\rho(x,z) + \rho(z,x)]^2 + \sigma^2(x,y)} \tag 2\\
\le &\ \sqrt{[\rho(x,z) + \rho(z,y)]^2 + [\sigma(x,z) + \sigma(z,y)]^2} \tag 3\\
\le &\ \sqrt{[\rho(x,z)]^2 + [\sigma(x,z)]^2} + \sqrt{[\rho(z,y)]^2 + [\sigma(z,y)]^2} \tag 4\\
= &\ \rho_2(x,z) + \rho_2(z,y).
\end{align}
$(2)$ and $(3)$ follow from the monotonicity we proved right above and the triangle inequality for the two metrics $\rho$ and $\sigma$. $(4)$ is a direct application of $(1)$.
