# Let $(X,d_{x}),(Y,d_{y})$ two metric spaces. Prove that $\sqrt{d_{x}(x_{1},x_{2})^{2}+d_{y}(y_{1},y_{2})^{2}}$ is a metric.

metric-spacesreal-analysis

d is defined by : $$d: (X\times Y)\times (X\times Y)\rightarrow \left [ 0, \infty \right ] : ((x_{1},y_{1}),(x_{2},y_{2})) \mapsto \sqrt{d_{x}(x_{1},x_{2})^{2}+d_{y}(y_{1},y_{2})^{2}}$$. I have proven everything but the triangle inequality condition which I'm asking for your help with. It also gives me the hint to use the Cauchy-Schwarz inequality in $$\mathbb{R}^{2}$$ which is: $$\forall a,b,c,d \in \mathbb{R}: ab+cd\leq \sqrt{a^{2}+c^{2}}\sqrt{b^{2}+d^{2}}$$.

In $$\mathbb{R}^2$$, $$\|(u,v)\|_2=\sqrt{u^2+v^2}$$ define a norm. In particular, $$\|(u+u',v+v')\|_2\leq\|(u,v)\|_2+\|(u',v')\|_2$$
With $$u=d_X(x,x'')$$, $$u'=d_X(x'',x)$$, $$v=d_Y(y,y'')$$ and $$v'=d_Y(y'',y)$$ \begin{align} d((x,y),(x',y')):&=\sqrt{d^2_X(x,x')+d^2_Y(y,y')}\leq\sqrt{(d_X(x,x'')+d_X(x'',x))^2+ (d_Y(y,y'')+d_Y(y'',y'))^2}\\ &=\|(u+u',v+v')\|_2\leq\|(u,v)\|_2+\|(u',v')\|_2\\ &=\sqrt{d^2_X(x,x'')+d^2_Y(y'',y)} + \sqrt{d^2_X(x'',x')+d^2_Y(y'',y')}\\ &=d((x,y),(x'',y'')) + d(x'',y''),(x',y')) \end{align}