trouble in getting triangle inequality Let $l_{2}$ be the set of all infinite sequences , $ (x_{n})$ such that $\sum_{n=1}^ {\infty} x_{n}$ converges.  Define $$d(x,y)= \sqrt{\sum_{n=1}^{\infty} (x_{n}-y_{n})^{2}}$$  for each $x=(x_{n})$ and $y=(y_{n}) \in l_{2}$.
I have to prove that $(l_{2},d)$ is a metric space.
Checking that it satisfies the first conditions of eing a metric is easy, and I've already done that, but it is the triangle inequality I'm having trouble with.  This is what I tried so far( most of which I think is wrong):
 $$ d(x,y) +d(y,z) = \sqrt{\sum_{n=1}^{\infty} (x_{n}-y_{n})^{2}} +\sqrt{\sum_{n=1}^{\infty} (y_{n}-z_{n})^{2}}$$
$$   \ge \sqrt{\sum_{n=1}^{\infty} \{(x_{n}-y_{n})^{2} +(y_{n}-z_{n})^{2}}\}    $$
because of the inequality that for $a \ge 0 , b\ge 0$ $ \sqrt{a}+\sqrt{b} \ge \sqrt{a+b}$
which is $$\ge \sqrt{\sum_{n=1}^{\infty} \{(x_{n}-y_{n}) +(y_{n}-z_{n})}\}$$   and it can esaily be seen that the last expression is equalt o $ d(x,z)$,.
Hence, $d(x,y)+ d(y,z) \ge d(x,z)$
but the process of obtaining the last step seems kind of incorrect to me.
Can somebody suggest a better way to get this inequality or if possible correct the method above?   
 A: Let's prove a version of the Minkowski inequality:
$$
\sqrt{\sum_{i=1}^na_i^2}+\sqrt{\sum_{i=1}^nb_i^2}\geq{\sqrt{\sum_{i=1}^n(a_i+b_i)^2}}\tag{$*$}.
$$
Squaring both sides and expanding the RHS, we see that (*) is equivalent to
$$
\sqrt{\sum_{i=1}^na_i^2}\sqrt{\sum_{i=1}^nb_i^2}\geq\sum_{i=1}^n a_ib_i
$$
which is true because it's just the Cauchy-Schwarz inequality. (You can prove it by noting that $0\leq |a-b\lambda|^2$ with $a=(a_1,\ldots,a_n)$ and $b=(b_1,\ldots,b_n)$ and considering the discriminant of $|a-b\lambda|^2$, which is a quadratic in $\lambda$.)
Apply (*) with $a_i=x_i-y_i$ and $b_i=y_i-z_i$, we have
$$
{\sqrt{\sum_{i=1}^n(z_i-x_i)^2}}\leq\sqrt{\sum_{i=1}^n(x_i-y_i)^2}+\sqrt{\sum_{i=1}^n(y_i-z_i)^2}\leq\sqrt{\sum_{i=1}^\infty(x_i-y_i)^2}+\sqrt{\sum_{i=1}^\infty(y_i-z_i)^2}
$$
which is $d(x,y)+d(y,z)$. Now the claim follows by letting $n\to\infty$ in the leftmost expression above to get $d(x,z)\leq d(x,y)+d(y,z)$.

In your approach, you wanted to say $\sqrt{\sum_{n=1}^{\infty}[(x_{n}-y_{n})^{2} +(y_{n}-z_{n})^{2}]}$ is greater than or equal to $\sqrt{\sum_{n=1}^{\infty} (x_n-z_{n})^{2}}$ presumably by term-by-term comparison. If so, then you would run into a problem because it's not necessary that $a_i^2+b_i^2\geq(a_i+b_i)^2$.
