I'm trying to show that $d(a,b) = \left( \sum_{n=1}^\infty|a_n - b_n|^2 \right)^{\frac{1}{2}}$ is a metric on the set of all sequences of real numbers $x = (x_1 , x_2, \dots , x_n , \dots )$ such that $\sum_{n=1}^\infty x_n^2$ converges.
However, I am having troubles showing it satisfies the triangle inequality.
$$d(a,c)\leq d(a,b)+d(b,c)$$
I attempted to start from $d(a,b)+d(b,c)$ and use $\sqrt{a}+\sqrt{b} \geq \sqrt{a+b}$ for $a$ and $b$ greater than zero as follows:
\begin{align} d(a,b) + d(b,c) &= \left( \sum_{n=1}^\infty|a_n - b_n|^2 \right)^{\frac{1}{2}} + \left( \sum_{n=1}^\infty|b_n - c_n|^2 \right)^{\frac{1}{2}}\\ &\geq \left( \sum_{n=1}^\infty|a_n - b_n|^2 + \sum_{n=1}^\infty|b_n - c_n|^2 \right)^{\frac{1}{2}}\\ &=\left( \sum_{n=1}^\infty\left(|a_n - b_n|^2 + |b_n - c_n|^2\right) \right)^{\frac{1}{2}} \end{align}
I must now prove $|a_n -b_n|^2 + |b_n - c_n|^2 \geq |a_n -c_n|^2$. Any hint or idea would be gladly appreciated.
