This is the most interesting proof—it depends on the equality $$
\sum_{i=1}^{n} x_i^2\cdot \sum_{i=1}^{n} y_i^2=\left( \sum_{i=1}^{n} x_iy_i \right) ^2+\sum_{i<j}\left( x_iy_j-x_jy_i \right) ^2
.$$ To check this equality, note that,
\begin{align*}
\sum_{i=1}^{n} x_i^2\cdot \sum_{i=1}^{n} y_i^2 &= \sum_{i=1}^{n} x_i^2y_i^2 +\sum_{i\neq j}x_i^2y_j^2 \\
\left( \sum_{i=1}^{n} x_iy_i \right) ^2 &= \sum_{i=1}^{n} \left( x_iy_i \right) ^2+\sum_{i\neq j}x_iy_ix_jy_j
\end{align*}
The difference is
\begin{align*}
\sum_{i\neq j}\left( x_i^2y_j^2-x_iy_ix_jy_j \right) &= 2\sum_{i<j}\left( x_i^2y_j^2+x_j^2y_i^2-x_iy_ix_jy_j \right) \\
&= 2\sum_{i<j}\left( x_iy_j-x_jy_i \right) ^2
\end{align*}
If equality holds in the Schwarz inequality, then all $x_iy_j=x_jy_i$. If some $y_i\neq 0$ say $y_1\neq 0$, then $x_i=\frac{x_1}{y_1}y_i$ for all $i$, so we can let $\lambda =\frac{x_1}{y_1}$.
So I got stuck trying to prove the Schwarz inequality by expressing the product of magnitudes as a sum of two squares, and was reading the solution from the answer book; however I could not understand the last part on "the difference", i.e. $ \sum_{i≠j}({x_i}^2{y_j}^2 – x_iy_ix_jy_j) = 2 \sum_{i<j} ({x_i}^2{y_j}^2 + {x_j}^2{y_i}^2 – x_iy_ix_jy_j) $. Please help me understand how the terms doubled and why the next line suddenly had a square term.
Best Answer
There are two typos in this page: On the first line after "The difference...", the "2" should be in front of $x_iy_ix_jy_j$, the reason being that when you change the index from $i\neq j$ to $i<j$ you have to double all the terms: say on the left hand side you have both $x_1^2y_2^2$ and $x_2^2y_1^2$, and since on the right hand side $I<j$ you have to include the second one. The same applies to the products $x_iy_ix_jy_j$. Each of the terms is obviously a square. There is a typo on the following line: there is no "2" in front of the sum, as in the original (and correct) statement of the inequality.