I have a doubt on how to prove the triangle inequality for the metric $d$ defined on $X\times X$ (where $X=X_{1}\times X_{2}$ and $(X_{1},d_{1}), (X_{2},d_{2})$ are metric spaces) by $$d(x,y)=\sqrt{d_{1}(x_{1},y_{1})^{2}+d_{2}(x_{2},y_{2})^{2}}$$ where $x=(x_{1},x_{2})$ and $y=(y_{1},y_{2})$.
I have tried to apply the triangle inequality for each metric but this produces a quantity "greater" than what i want. Any sugestions?
For convenience let $a=d_1(x_1,y_1)$, $b=d_1(x_1,z_1)$, $c=d_1(z_1,y_1)$, $u=d_2(x_2,y_2)$, $v=d_2(x_2,z_2)$, and $w=d_2(z_2,y_2)$; we know that $a\le b+c$ and $u\le v+w$, and we want to show that
$$\sqrt{a^2+u^2}\le\sqrt{b^2+v^2}+\sqrt{c^2+w^2}\,.\tag{1}$$
Lacking any brilliant ideas, we might do the obvious and square both sides, noting that we’ll have $(1)$ if we can show that
$$a^2+u^2\le b^2+v^2+c^2+w^2+2\sqrt{(b^2+v^2)(c^2+w^2)}\,.$$
We do know that $a^2\le b^2+c^2+2bc$ and $u^2\le v^2+w^2+2vw$, so
$$a^2+u^2\le b^2+c^2+v^2+w^2+2(bc+vw)\,,$$
and we’ll be done if we can show that
$$bc+vw\le\sqrt{(b^2+v^2)(c^2+w^2)}\,.$$
We’re still dealing with non-negative quantities, so we can square again and hope to show that
$$(bc+vw)^2\le(b^2+v^2)(c^2+w^2)\,,$$
i.e., that
$$b^2c^2+v^2w^2+2bcvw\le b^2c^2+v^2w^2+b^2w^2+v^2c^2\,.$$
And that reduces to showing that $2bcvw\le b^2w^2+v^2c^2$, which is easily done by a standard trick:
$$0\le(bw-vc)^2=b^2w^2+v^2c^2-2bcvw\,,$$
so $2bcvw\le b^2w^2+v^2c^2$.
This is very inelegant and could doubtless be done much more neatly, but it shows that a bit of intelligently directed tinkering can sometimes be good enough even if you don’t at first see just where you’re going.
