Cauchy-Schwarz Inequality I have to show the following is equivalent to the Cauchy-Schwarz Inequality:
$$\left( \sum (r_i +s_i )^2 \right)^{\frac {1}{2}} \leq \left( \sum r_i ^2 \right)^{\frac {1}{2}}  + \left( \sum s_i ^2 \right)^{\frac {1}{2}}$$
The Cauchy-Schwarz Inequality is defined as :
$$ \left( \sum {(r_i s_i)} \right)^2 \leq  \left( \sum{r_i}^2 \right)\left( \sum{s_i}^2 \right) $$
My Attempt:
Since both sides are non-negative we can square both sides.
$$ \sum (r_i +s_i )^2 \leq \sum {r_i ^2}  +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} +  \sum s_i ^2  $$
Expanding the left-hand side.
$$ \sum (r_i^2 + 2r_is_i + s_i^2 )\leq \sum {r_i ^2}  +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} +  \sum s_i ^2  $$
Using identities of Summation.
$$ \sum r_i^2 + 2\sum {(r_i s_i)} + \sum s_i^2 \leq \sum {r_i ^2}  +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} +  \sum s_i ^2  $$
Cancelling like terms 
$$
 \require{cancel}
\cancel  {\sum r_i^2} + 2\sum {(r_i s_i)} + \cancel{\sum s_i^2} \leq \cancel{\sum {r_i ^2}}  +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} +  \cancel{\sum s_i ^2}  
$$
We obtain the following:
$$ 2\sum {(r_i s_i)}  \leq  2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}}
$$
Dividing by 2:
$$ \sum {(r_i s_i)}  \leq   \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}}
$$
It is at this point that I am stuck. Does squaring both sides preserve the inequality at this point?
Thanks in advance.
 A: 
Does squaring both sides preserve the inequality at this point?

In a word, "yes":
$$
\sum(r_{i} s_{i})
  \leq \left|\sum(r_{i} s_{i})\right|
  \leq \left(\sum r_{i}^{2}\right)^{1/2} \left(\sum s_{i}^{2}\right)^{1/2},
$$
with the second inequality being the square root of "your" Cauchy-Schwarz inequality.

As a stylistic point, two-column proofs are a (less-than-optimal) habit from school courses that should generally be outgrown if you aspire to be a mathematician. In published work, mathematicians instead build chains of equalities and inequalities. Here, for example, you might argue that "your" C-S inequality implies the "new" one like this:
\begin{align*}
\sum (r_{i} + s_{i})^{2}
  &= \sum r_{i}^{2} + 2 \sum (r_{i}s_{i}) + \sum s_{i}^{2} \\
  &\leq \sum r_{i}^{2} + 2 \left|\sum (r_{i}s_{i})\right| + \sum s_{i}^{2} \\
  &\leq \sum r_{i}^{2} + 2 \left(\sum r_{i}^{2}\right)^{1/2} \left(\sum s_{i}^{2}\right)^{1/2} + \sum s_{i}^{2} \\
  &= \left[\left(\sum r_{i}^{2}\right)^{1/2} + \left(\sum s_{i}^{2}\right)^{1/2}\right]^{2}.
\end{align*}
