Michael Spivak's Calculus - Chapter 1, Problem 19 Problem 19. The fact that ${a^2}\ge{0}$ for all the numbers a, elementary as it may seem, is nevertheless the fundamental idea upon which most important inequalities are ultimately based. The great-grandaddy of all inequalities is the Schwarz inequality:
${x_{1}y_{1}+x_{2}y_{2}}\le{\sqrt{x_{1}^2+x_{2}^2}\sqrt{y_{1}^2+y_{2}^2}}$
The three proofs of the Schwarz inequality outlineed below have only one thing in common - their reliance on the fact that $a^2\ge{0}$ for all $a$.
(a) Using Problem 18, complete the proof of the Schwarz inequality.
Solution.
This is the proof I've written.
\begin{array}{ll}
(x_{1}y_{2}-x_{2}y_{1})^{2} & \ge0\\
x_{1}^{2}y_{2}^{2}-2x_{1}x_{2}y_{1}y_{2}+x_{2}^{2}y_{1}^{2} & \ge0\\
x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2} & \ge2x_{1}x_{2}y_{1}y_{2}\\
2x_{1}x_{2}y_{1}y_{2} & \le x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2}\\
x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}+2x_{1}x_{2}y_{1}y_{2} & \le x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2}\\
(x_{1}y_{1}+x_{2}y_{2})^{2} & \le(x_{1}^{2}+x_{2}^{2})(y_{1}^{2}+y_{2}^{2})\\
x_{1}y_{1}+x_{2}y_{2} & \le\sqrt{(x_{1}^{2}+x_{2}^{2})}\sqrt{(y_{1}^{2}+y_{2}^{2})}
\end{array}
(b)Prove the Schwarz inequality by using $2xy\le{x^2+y^2}$ with
$x=\displaystyle{\frac{x_{i}}{\sqrt{x_{1}^2+x_{2}^2}}}$, $y=\displaystyle{\frac{y_{i}}{\sqrt{x_{1}^2+x_{2}^2}}}$
first for $i=1$ and then for $i=2$. 
I am not able use the inequality to come up with another proof. Could someone tell me the way, how to approach this. 
Also, what are some of the applications of Cauchy-Schwarz inequality? I did google it, but would love to hear it from a mathematician!
 A: Here is the argument Spivak provides in his Combined Answer Book — (for 19a):
Supposing $y_1$ and $y_2$ are not both $0$, and that there is no number $\lambda$ such that $x_1=\lambda y_1$ and $x_2=\lambda y_2$, then $$\begin{array}{tcl}0 &<& (\lambda y_1-x_1)^2 + (\lambda y_2-x_2)^2 \\ &=& \lambda^2 (y_1^2+y_2^2)-2\lambda(x_1y_1+x_2y_2)+(x_1^2+x_2^2),\end{array}$$ and the equation $$\lambda^2 (y_1^2+y_2^2)-2\lambda(x_1y_1+x_2y_2)+(x_1^2+y_1^{2 \ *})=0 \\$$ 
has no solution $\lambda$. So by problem 18(a) we must have $$\Bigg[\frac{2(x_1y_1+x_2y_2)}{({y_1}^1+{y_2}^2)}\Bigg]^2-\frac{4({x_1}^2+{y_1}^{2 \ *})}{({y_1}^2+{y_2}^2)} < 0, \\$$ which yields the Schwarz inequality. ( * It is possible that this term was a typo, and intended to be $"x_2^2"$ – as the line that contains it is simply an expansion of the right–hand side of the initial inequality, and otherwise identical to the equation just above it. Given 18(a), the argument seems erroneous otherwise. But the simplification actually works out using either term, so I may be missing some other point. Feel free to comment below.)
A: We have
\begin{align}2x_iy_i &\le \frac{x_i^2 \sqrt{y_1^2 + y_2^2}}{\sqrt{x_1^2 + x_2^2}} + \frac{y_i^2 \sqrt{x_1^2 + x_2^2}}{\sqrt{y_1^2 + y_2^2}}\\
&\le \frac{x_i^2 (y_1^2 + y_2^2) + y_i^2 (x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
&\le \frac{x_i^2 (y_1^2 + y_2^2) + y_i^2 (x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
\to 2 \sum_{i\in\{1,2\}} x_iy_i &\le \frac{(x_1^2 + x_2^2) (y_1^2 + y_2^2) + (y_1^2 + y_2^2) (x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
\to \sum_{i\in\{1,2\}} x_iy_i &\le \frac{(x_1^2 + x_2^2) (y_1^2 + y_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
\to \sum_{i\in\{1,2\}} x_iy_i &\le \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}\end{align}
QED
A: The general case of Cauchy-Schwarz inequality can be proved by Lagrange Identity, as in Prove Lagrange's Identity without induction.
