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.