I'm trying to prove that $\sum_{j=1}^n\langle x_j,y_j\rangle \leq \sqrt{ \sum_{j=1}^n\langle x_j,x_j\rangle} \sqrt{\sum_{j=1}^n \langle y_j,y_j\rangle}$, where $\left(V, \langle \cdot , \cdot \rangle \right)$ is an inner product space and $x_1, x_2, ...$ and $y_1, y_2,...$ are sequences of elements of $V$.
My attempt: $$\sum_{j=1}^n\langle x_j,y_j\rangle \leq \sum_{j=1}^n\langle x_j,x_j\rangle^{1/2}\langle y_j,y_j\rangle^{1/2} \leq \left(\sum_{j=1}^n \langle x_j,x_j\rangle\right)^{1/2} \left(\sum_{j=1}^n \langle y_j,y_j\rangle\right)^{1/2}.$$
I'm not sure if the first inequality holds for an arbitrary inner product space. The second inequality is based on Cauchy. Am I correct?
I probably write it as the answer, I think it would be clearer this way.
So, you have some scalar product $\langle,\rangle$ on $V\times V$. As with every scalar product, you have the Cauchy inequality: $$|\langle x,y\rangle| \le \langle x,x\rangle^{1/2}\langle y,y\rangle^{1/2}.$$
It is valid for all possible $x,y$, so also for $x_i,y_i$. Hence you have for each $i$: $$|\langle x_i,y_i\rangle| \le \langle x_i,x_i\rangle^{1/2}\langle y_i,y_i\rangle^{1/2}.$$ Summing up you then have $\sum_{i=1}^n|\langle x_i,y_i\rangle| \le \sum_{i=1}^n\langle x_i,x_i\rangle^{1/2}\langle y_i,y_i\rangle^{1/2}$.
In the next step you consider another space: $\Bbb R^n$. In this space there is the standard scalar product and then the Cauchy inequality with respect to this product. It says that for each $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ you have $$\sum_{k=1}^n |x_iy_i|\le \left(\sum_{i=1}^n |x_i|^2\right)^{1/2} \left(\sum_{i=1}^n |y_i|^2\right)^{1/2}.$$
As it holds for all $x,y$, it also holds for $x=(\langle x_1,x_1\rangle^{1/2},...,\langle x_n,x_n\rangle^{1/2})$ and $y=(\langle y_1,y_1\rangle^{1/2},...,\langle y_n,y_n\rangle^{1/2})$ which are simply the vectors from $\Bbb R^n$. Now for these concrete vectors, Cauchy says that $$\sum_{i=1}^n \langle x_i,x_i\rangle^{1/2} \langle y_i,y_i\rangle^{1/2}\le \left(\sum_{i=1}^n \langle x_i,x_i\rangle\right)^{1/2} \left(\sum_{i=1}^n \langle y_i,y_i\rangle\right)^{1/2}.$$ This proves the second inequality.