Let $\{x_n\}$ and $\{y_n\}$ are real number sequences.
Proof of this inequality
$$
\sqrt{
\sum_{n=1}^{k} (x_n-y_n)^2}
\leq
\sqrt{
\sum_{n=1}^{k} x_n^2}
+
\sqrt{
\sum_{n=1}^{k} y_n^2}$$
My mathematics book says that we can prove this inequality by using Cauchy–Schwarz inequality, but I don't know how I have to use Cauchy–Schwarz inequality.
Does anyone know how to use Cauchy–Schwarz inequality in this situation?
Best Answer
By C-S $$\sqrt{\sum_{n=1}^kx_n^2}+\sqrt{\sum_{n=1}^ky_n^2}=\sqrt{\left(\sqrt{\sum_{n=1}^kx_n^2}+\sqrt{\sum_{n=1}^ky_n^2}\right)^2}=$$ $$=\sqrt{\sum_{n=1}(x_n^2+y_n^2)+2\sqrt{\sum_{n=1}^kx_n^2\sum_{n=1}^ky_n^2}}\geq\sqrt{\sum_{n=1}^k(x_n^2+y_n^2)+2\sqrt{\left(\sum_{n=1}^kx_ny_n\right)^2}}=$$ $$=\sqrt{\sum_{n=1}^k(x_n^2+y_n^2)+2\left|\sum_{n=1}^kx_ny_n\right|}\geq\sqrt{\sum_{n=1}^k(x_n^2+y_n^2)-2\sum_{n=1}^kx_ny_n}=\sqrt{\sum_{n=1}^k(x_n-y_n)^2}.$$ Also, we can use Minkowski here: $$\sqrt{\sum_{n=1}^kx_n^2}+\sqrt{\sum_{n=1}^ky_n^2}=\sqrt{\sum_{n=1}^kx_n^2}+\sqrt{\sum_{n=1}^k(-y_n)^2}\geq\sqrt{\sum_{n=1}^k(x_n-y_n)^2}.$$